Limit Fungsi Trigonometri

Tujuan Pembelajaran:
Menentukan nilai limit fungsi trigonometri.

Konsep Limit

Ingat pengertian limit seperti pada https://kelasadem.github.io/matematika-sma/limit/.
Di sini ditulis ulang pengertian limit secara intuitif.

Pengertian limit secara intuitif

\(\lim\limits_{x \to c} f(x) = L\) berarti nilai \(L\) adalah nilai \(f(x)\), jika \(x\) dekat ke \(c\) (tidak berarti \(x\) sama dengan \(c\), melainkan \(x\) dekat ke \(c\) dari arah kiri dan \(x\) dekat ke \(c\) dari arah kanan).

Jika fungsi \(f(x)\) terdefinisi pada interval yang memuat \(x=c\), maka:

  1. \(\lim\limits_{x \to c} f(x) = L\) jika dan hanya jika \(\lim\limits_{x \to c^-} f(x) = L\) dan \(\lim\limits_{x \to c^+} f(x) = L\)
  2. Jika \(\lim\limits_{x \to c^-} f(x) = L_1\) dan \(\lim\limits_{x \to c^+} f(x) = L_2\) dengan \(L_1 \neq L_2\), maka \(\lim\limits_{x \to c} f(x)\) tidak ada.

Keterangan

  • \(\lim\limits_{x \to c^-} f(x)\) dibaca limit \(f(x)\) untuk nilai \(x\) yang mendekati \(c\) dari kiri (\(x<c\)).
  • \(\lim\limits_{x \to c^+} f(x)\) dibaca limit \(f(x)\) untuk nilai \(x\) yang mendekati \(c\) dari kanan (\(x>c\)).

Kita akan meninjau nilai \(f(x)=\dfrac{\sin{x}}{x}\) di sekitar \(x=0\), yaitu menentukan nilai \(\lim\limits_{x\to{0}}\dfrac{\sin{x}}{x}\).

Perhatikan tabel berikut.

\(x\) -0,1 -0,01 -0,001 0 0,001 0,01 0,1
\(f(x)=\dfrac{\sin{x}}{x}\) 0,99833 0,99998 0,9999998 ? 0,9999998 0,99998 0,99833

Kita dapat melihat bahwa:

  • saat \(x\) mendekati 0 dari arah kiri, nilai \(\dfrac{\sin{x}}{x}\) mendekati 1, kita tulis \(\lim\limits_{x\to{0^-}}\dfrac{\sin{x}}{x}=1\). Ini kita sebut limit kirinya.
  • saat \(x\) mendekati 0 dari arah kanan, nilai \(\dfrac{\sin{x}}{x}\) mendekati 1 juga, kita tulis \(\lim\limits_{x\to{0^+}}\dfrac{\sin{x}}{x}=1\). Ini kita sebut limit kanannya.
  • Karena limit kiri dan limit kanan bernilai sama, maka dapat kita simpulkan bahwa \(\lim\limits_{x\to{0}}\dfrac{\sin{x}}{x}=1\)

Teorema Limit

Teorema Utama Limit

Jika \(I\) adalah interval nilai \(x\in\mathbb{R}\) dengan \(a<x<b\), \(c\in I\), \(k\in\mathbb{R}\), \(n\in\mathbb{Z^+}\), serta \(\lim\limits_{x \to c} f(x)\), \(\lim\limits_{x \to c} g(x)\), dan \(\lim\limits_{x \to c} h(x)\) terdefinisi, maka berlaku:

  1. \(\lim\limits_{x \to c} k = k\)
  2. \(\lim\limits_{x \to c} x = c\)
  3. \(\lim\limits_{x \to c} k \cdot f(x) = k \times \lim\limits_{x \to c} f(x)\)
  4. \(\lim\limits_{x \to c} \left( f(x) \pm g(x) \right) = \lim\limits_{x \to c} f(x) \pm \lim\limits_{x \to c} g(x)\)
  5. \(\lim\limits_{x \to c} \left( f(x) \times g(x) \right) = \lim\limits_{x \to c} f(x) \times \lim\limits_{x \to c} g(x)\)
  6. \(\lim\limits_{x \to c} \dfrac{f(x)}{g(x)} = \dfrac{\lim\limits_{x \to c} f(x)}{\lim\limits_{x \to c} g(x)}\) dengan \(\lim\limits_{x \to c} g(x) \neq 0\)
  7. \(\lim\limits_{x \to c} \left( f(x) \right)^n = \left( \lim\limits_{x \to c} f(x) \right)^n\)
  8. \(\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{\lim\limits_{x \to c} f(x)}\) dengan \(\lim\limits_{x\to c} f(x) > 0\) untuk \(n\) genap.

Contoh 1

Ingat:
\(\lim\limits_{x \to c} k = k\)

Maka

\(\lim\limits_{x \to 0} \pi = \pi\)

Ingat:
\(\lim\limits_{x \to c} x = c\)

Maka

\(\lim\limits_{x \to \frac{1}{4}\pi} x = \frac{1}{4}\pi\)

Ingat:
\(\lim\limits_{x \to c} k\cdot f(x) = k \times \lim\limits_{x \to c} f(x)\)

Maka

\[\begin{align*}\lim\limits_{x \to \frac{1}{6}\pi} 8\sin{x} &= 8\times \lim\limits_{x \to \frac{1}{6}\pi} \sin{x} \\ &= 8\times \sin{\frac{1}{6}\pi} \\ &= 8\times \frac{1}{2} \\ &= 4\end{align*}\]

Ingat:
\(\lim\limits_{x \to c} \left( f(x) \pm g(x) \right) = \lim\limits_{x \to c} f(x) \pm \lim\limits_{x \to c} g(x)\)

Maka

\[\begin{align*} \lim\limits_{x \to \frac{1}{3}\pi} \left( \sin{x}+\cos{x}-\tan{x} \right) &= \lim\limits_{x \to \frac{1}{3}\pi} \sin{x} + \lim\limits_{x \to \frac{1}{3}\pi} \cos{x} - \lim\limits_{x \to \frac{1}{3}\pi} \tan{x} \\ &= \sin{\frac{1}{3}\pi} + \cos{\frac{1}{3}\pi} - \tan{\frac{1}{3}\pi} \\ &= \frac{1}{2}\sqrt{3} + \frac{1}{2} -\sqrt{3} \\ &= \frac{1}{2} - \frac{1}{2}\sqrt{3} \\ &= \frac{1}{2}\left(1-\sqrt{3}\right) \end{align*}\]

Ingat:
\(\lim\limits_{x \to c} \left( f(x) \times g(x) \right) = \lim\limits_{x \to a} f(x) \times \lim\limits_{x \to a} g(x)\)

Maka

\[\begin{align*} &\lim\limits_{x \to \frac{1}{4}\pi} \sin{x}\times\cos{x} \\ &= \lim\limits_{x \to \frac{1}{4}\pi} \sin{x} \times \lim\limits_{x \to \frac{1}{4}\pi} \cos{x} \\ &= \sin{\frac{1}{4}}\times\cos{\frac{1}{4}} \\ &= \frac{1}{2}\sqrt{2}\times\frac{1}{2}\sqrt{2} \\ &= \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4}=\frac{1}{2} \end{align*}\]

Ingat:
\(\lim\limits_{x \to c} \dfrac{f(x)}{g(x)} = \dfrac{\lim\limits_{x \to c} f(x)}{\lim\limits_{x \to c} g(x)}\) dengan \(\lim\limits_{x \to c} g(x) \neq 0\)

Maka

\[\begin{align*} \lim\limits_{x \to 0} \frac{1-2\,\sin{x}}{\cos{x}} &= \frac{\lim\limits_{x \to 0} (1-2\,\sin{x})}{\lim\limits_{x \to 0} \cos{x}} \\ &= \frac{1-2\cdot{0}}{1} = \frac{1}{1}=1 \end{align*}\]

Ingat:
\(\lim\limits_{x \to c} \left( f(x) \right)^n = \left( \lim\limits_{x \to c} f(x) \right)^n\)

Maka

\[\begin{align*} \lim\limits_{x \to \frac{1}{6}\pi} \left(1-\sin^2{x}\right)^{3} &= \lim\limits_{x \to \frac{1}{6}\pi} \left(\cos^2{x}\right)^{3} \\ &= \lim\limits_{x \to \frac{1}{6}\pi} \left(\cos{x}\right)^{2\times{3}} \\ &= \left(\lim\limits_{x \to \frac{1}{6}\pi}\cos{x}\right)^{6} \\ &= \left(\frac{1}{2}\sqrt{3}\right)^{6} \\ &= \frac{27}{64} \end{align*}\]

Ingat:
\(\lim\limits_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{\lim\limits_{x \to c} f(x)}\)

Maka

\[\begin{align*} \lim\limits_{x \to \frac{1}{3}\pi} \sqrt{\sin^2{x}+\cos{x}} &= \sqrt{\lim\limits_{x \to \frac{1}{3}\pi} (\sin^2{x}+\cos{x})} \\ &= \sqrt{\lim\limits_{x \to \frac{1}{3}\pi}\sin^2{x} + \lim\limits_{x \to \frac{1}{3}\pi} \cos{x})} \\ &= \sqrt{\left(\lim\limits_{x \to \frac{1}{3}\pi}\sin{x}\right)^2 + \frac{1}{2}} \\ &= \sqrt{\left(\frac{1}{2}\sqrt{3}\right)^2 + \frac{1}{2}} \\ &= \sqrt{\frac{3}{4} + \frac{2}{4}} = \sqrt{\frac{5}{4}} = \frac{1}{2}\,\sqrt{5} \end{align*}\]

Hukum substitusi:

Jika \(f(x)\) kontinu seperti:

  1. \(f(x)\) fungsi polinomial, maka: \[\lim\limits_{x \to c} f(x) = f(c)\]
  2. \(f(x)\) fungsi rasional, \(f(x)=\dfrac{g(x)}{h(x)}\) dan \(f(c)\) terdefinisi, maka: \[\lim\limits_{x \to c} f(x) = f(c)=\frac{g(c)}{h(c)}\]

Contoh 2

Tentukan nilai \(\lim\limits_{x \to \pi} \cos{x}(1-\sin{x})\).

Jawab

Karena fungsi \(f(x)=\cos{x}(1-\sin{x})\) kontinu pada \(x\in\mathbb{R}\), maka:

\[\begin{align*}\lim\limits_{x \to \pi} \cos{x}(1-\sin{x}) &= \cos{\pi}(1-\sin{\pi}) \\ &= 1\times(1-0) \\ &= 1 \end{align*}\]

Tentukan nilai \(\lim\limits_{x \to 0} \dfrac{\cos{x}}{1-\sin{x}}\).

Jawab

Karena fungsi \(f(x)=\dfrac{\cos{x}}{1-\sin{x}}\) kontinu pada saat nilai \(x\) di sekitar \(0\), maka:

\[\begin{align*}\lim\limits_{x \to 0} \dfrac{\cos{x}}{1-\sin{x}} &= \frac{\cos{0}}{1-\sin{0}} \\ &= \frac{1}{1-0} = 1 \end{align*}\]

Teorema Apit:

Untuk \(I\) adalah interval nilai \(x\in\mathbb{R}\) dengan \(a<x<b\), \(c\in I\), dan \(g(x) \leq f(x) \leq h(x)\) untuk setiap \(x \in I\), jika \(\lim\limits_{x \to c} g(x) = L\) dan \(\lim\limits_{x \to c} h(x) = L\), maka \(\lim\limits_{x \to c} f(x) = L\)

Teorema Apit biasa disebut juga Sequeeze Teorem atau Teorema Sandwich.

Contoh 3

Buktikan bahwa \(\lim\limits_{x \to 0} \dfrac{\sin{x}}{x}=1\)

Jawab:

Berikut contoh penggunaan Teorema Apit dalam membuktikan bahwa: \[\lim\limits_{x \to 0}\frac{\sin{x}}{x}=1\]

Perhatikan lingkaran berjari-jari 1 dan lingkaran berjari-jari \(\cos{\theta}\) serta luas juring OBQ, luas segitiga OBP, dan luas juring OAP pada gambar berikut.

Luas juring suatu lingkaran dengan sudut pusat \(\theta\) adalah \(\dfrac{\theta}{2\pi}\times\text{Luas lingkaran}\).

Maka:

  • Luas juring OBQ = \(\frac{\theta}{2\pi}\times{\pi}\times(\cos{\theta})^2 = \frac{1}{2}\,\theta\,\cos^2{\theta}\)
    Dapat kita lihat bahwa nilai \(\theta\,\cos{\theta}>0\) (positif)
  • Luas segitiga OBP = \(\frac{1}{2}\times \cos{\theta} \times \sin{\theta}\)
  • Luas juring OAP = \(\frac{\theta}{2\pi}\times\pi\times(1)^2 = \frac{1}{2}\,\theta\)

Luas segitiga OBP diapit oleh (bernilai di antara) luas juring OBQ dan luas juring OAP.

Sehingga diperoleh:

\(L_\text{Juring OBQ} \leq L_{\triangle\text{OBP}} \leq L_{\text{Juring OAP}}\)

\(\Leftrightarrow \frac{1}{2}\,\theta\cdot\cos^2{\theta} \leq \frac{1}{2}\,\cos{\theta}\cdot\sin{\theta} \leq \frac{1}{2}\,\theta\) \(\small\color{cadetblue}\qquad\times\frac{2}{\theta\,\cos{\theta}}\)

\(\Leftrightarrow \cos{\theta} \leq \dfrac{\sin{\theta}}{\theta} \leq \dfrac{1}{\cos{\theta}}\)

\(\Leftrightarrow \lim\limits_{\theta \to 0}\cos{\theta} \leq \lim\limits_{\theta \to 0}\dfrac{\sin{\theta}}{\theta} \leq \lim\limits_{\theta \to 0}\dfrac{1}{\cos{\theta}}\)

\(\Leftrightarrow 1 \leq \lim\limits_{\theta \to 0} \dfrac{\sin{\theta}}{\theta} \leq 1\)

Karena nilai \(\lim\limits_{\theta\to 0}\dfrac{\sin{\theta}}{\theta}\) diapit oleh nilai 1 dan 1, maka \[\lim\limits_{\theta\to 0}\frac{\sin{\theta}}{\theta}=1\]
∎

Sifat-sifat Limit (1)

  1. \(\lim\limits_{x\to{c}}\sin{x}=\sin{c}\,\) dan \(\,\lim\limits_{x\to{c}}\cos{x}=\cos{c}\)

  2. \(\lim\limits_{x\to{0}}\dfrac{\sin{x}}{x}=1\,\) dan \(\,\lim\limits_{x\to{0}}\dfrac{x}{\sin{x}}=1\)

  3. \(\lim\limits_{x\to{0}}\dfrac{\tan{x}}{x}=1\,\) dan \(\,\lim\limits_{x\to{0}}\dfrac{x}{\tan{x}}=1\)

Contoh 4

4.1. Tentukan nilai \(\lim\limits_{x \to 0} (\cos{x}-\sin{x})\)

Jawab:

Karena fungsi \(f(x)=\cos{x}-\sin{x}\) kontinu pada \(x\in\mathbb{R}\), maka:

\(\begin{flalign*}\lim\limits_{x \to 0} \cos{x}-\sin{x} &= \cos{0}-\sin{0} \\ &= 1-0 \\ &= 1 \end{flalign*}\)

Jawab:

Pada Contoh 3 sudah dibuktikan bahwa \(\lim\limits_{x\to{0}}\dfrac{\sin{x}}{x}=1\).

Sekarang, akan dibuktikan bahwa: \[\lim\limits_{x \to 0}\frac{x}{\sin{x}}=1\]

Perhatikan lingkaran berpusat di \(O\), berjari-jari \(r\), melalui titik \(A\) dan \(B\) serta terbentuk segitiga siku-siku \(OCB\), juring \(OAB\), dan segitiga siku-siku \(OAD\) pada gambar berikut.

Dapat dihitung bahwa:

\(\begin{flalign*}\text{Luas }\triangle OCB &= \frac{1}{2} \times OC \times CB \\ &= \frac{1}{2} \times r \times \cos{x} \times r \times \sin{x} \\ &= \frac{1}{2}\, r^2\, \cos{x} \, \sin{x} \end{flalign*}\)

\(\begin{flalign*}\text{Luas juring } OAB &= \frac{x}{2\pi} \times \pi \cdot r^2 \\ &= \frac{1}{2}\, x \, r^2 \end{flalign*}\)

\(\begin{flalign*}\text{Luas }\triangle OAD &= \frac{1}{2} \times OA \times AD \\ &= \frac{1}{2} \times r \times r \times \tan{x} \\ &= \frac{1}{2} \, r^2 \, \frac{\sin{x}}{\cos{x}} \end{flalign*}\)

Luas juring \(OAB\) diapit oleh (bernilai di antara) luas \(\triangle OCB\) dan luas \(\triangle OAD\).

Sehingga diperoleh:

\(L_{\triangle OCB} \leq L_{\text{juring }OAB} \leq L_{\triangle{OAD}}\)

\(\Leftrightarrow \frac{1}{2}\,r^2\,\cos{x}\,\sin{x} \leq \frac{1}{2}\, x\, r^2 \leq \frac{1}{2}\,r^2\,\dfrac{\sin{x}}{\cos{x}}\) \(\small\color{cadetblue}\qquad\times\frac{2}{r^2\,\sin{x}}\)

\(\Leftrightarrow \cos{x} \leq \dfrac{x}{\sin{x}} \leq \dfrac{1}{\cos{x}}\)

\(\Leftrightarrow \lim\limits_{x \to 0}\cos{x} \leq \lim\limits_{x\to 0}\dfrac{x}{\sin{x}} \leq \lim\limits_{\theta \to 0}\dfrac{1}{\cos{x}}\)

\(\Leftrightarrow 1 \leq \lim\limits_{x\to{0}} \dfrac{x}{\sin{x}} \leq 1\)

Karena nilai \(\lim\limits_{x\to{0}}\dfrac{x}{\sin{x}}\) diapit oleh nilai 1 dan 1, maka \[\lim\limits_{x\to{0}}\frac{x}{\sin{x}}=1\]
∎
4.2b. Tentukan nilai \(\lim\limits_{x \to 0} \dfrac{\sin{2x}}{4x}\)

Jawab:

\(\begin{flalign*} \lim\limits_{x \to 0} \frac{\sin{2x}}{4x} &= \lim\limits_{x \to 0} \frac{\sin{2x}}{2 \times 2x} \qquad x\to{0}\,\Rightarrow\, 2x\to{0}\\ &= \frac{1}{2}\times\lim\limits_{2x \to 0} \frac{\sin{2x}}{2x} \\ &= \frac{1}{2}\times{1}=\frac{1}{2} \end{flalign*}\)

4.2c. Tentukan nilai \(\lim\limits_{x \to 0} \dfrac{\sin^2{8x}}{2x\,\tan{4x}}\)

Jawab: Ingat \(x\to{0}\,\Leftrightarrow\, 4x\to{0}\,\Leftrightarrow\, 8x\to{0}\)

\(\begin{flalign*} \lim\limits_{x \to 0} \frac{\sin^2{8x}}{2x\,\tan{4x}} &= \lim\limits_{x \to 0} \frac{\sin{8x}\times\sin{8x}}{{\color{orange}2}x\,\tan{4x}}\times\frac{{\color{red}8}\times{\color{green}8}\times{\color{blue}4}\times{\color{brown}x}}{{\color{red}8}\times{\color{green}8}\times{\color{blue}4}\times{\color{brown}x}}\\ &= \frac{{\color{red}8}\times{\color{green}8}}{{\color{orange}2}\times{\color{blue}4}}\times\lim\limits_{8x \to 0} \frac{\sin{8x}}{{\color{red}8}x}\times\lim\limits_{8x \to 0}\frac{\sin{8x}}{{\color{green}8}{\color{brown}x}}\times\lim\limits_{4x \to 0}\frac{{\color{blue}4}{\color{brown}x}}{\tan{4x}} \\ &= 8\times{1}\times{1}\times{1}=8 \end{flalign*}\)

4.3. Tentukan nilai dari \(\lim\limits_{x\to{0}}\dfrac{\sin{x}\cdot\tan{x}}{x^2}\)

Jawab:

\(\begin{flalign*} \lim\limits_{x\to{0}}\dfrac{\sin{x}\cdot\tan{x}}{x^2} &= \lim\limits_{x\to{0}}\dfrac{\sin{x}}{x} \times \lim\limits_{x\to{0}}\dfrac{\tan{x}}{x} \\ &= 1 \times 1 = 1 \end{flalign*}\)

Sifat-sifat Limit (2)

  1. \(\lim\limits_{x\to{0}}\dfrac{\sin{ax}}{bx}=\frac{a}{b}\,\) dan \(\,\lim\limits_{x\to{0}}\dfrac{ax}{\sin{bx}}=\frac{a}{b}\)

  2. \(\lim\limits_{x\to{0}}\dfrac{\tan{ax}}{bx}=\frac{a}{b}\,\) dan \(\,\lim\limits_{x\to{0}}\dfrac{ax}{\tan{bx}}=\frac{a}{b}\)

  3. \(\lim\limits_{x\to{0}}\dfrac{\sin{ax}}{\tan{bx}}=\frac{a}{b}\,\) dan \(\,\lim\limits_{x\to{0}}\dfrac{\tan{ax}}{\sin{bx}}=\frac{a}{b}\)

Contoh 5

5a. Tentukan nilai \(\lim\limits_{x \to 0} \dfrac{\sin^2{6x}}{2x\,\tan{9x}}\)

Jawab:

\(\begin{flalign*} \lim\limits_{x \to 0} \frac{\sin^2{6x}}{2x\,\tan{9x}} &= \lim\limits_{x \to 0} \frac{\sin{6x}\times\sin{6x}}{2x\,\tan{9x}} \\ &= \lim\limits_{x \to 0} \frac{\sin{6x}}{2x} \times \lim\limits_{x \to 0} \frac{\sin{6x}}{\tan{9x}} \\ &= \frac{6}{2} \times \frac{6}{9} \\ &= 3 \times \frac{2}{3} = 2 \end{flalign*}\)

5b. Tentukan nilai \(\lim\limits_{x \to 0} \dfrac{\tan^2{6x}}{\sin{6x}}\)

Jawab:

\(\begin{flalign*} \lim\limits_{x \to 0} \frac{\tan^2{6x}}{\sin{6x}} &= \lim\limits_{x \to 0} \frac{\tan{6x}\times\tan{6x}}{\sin{6x}} \\ &= \lim\limits_{x \to 0} \frac{\tan{6x}}{\sin{6x}} \times \lim\limits_{x \to 0} \tan{6x} \\ &= 1 \times \tan{(6\times{0})} \\ &= 1 \times \tan{0} = 1 \times 0 = 0 \end{flalign*}\)

5c. Tentukan nilai \(\lim\limits_{x \to 0} \dfrac{\sin{x}+\tan{3x}-\sin{5x}}{x-\sin{2x}-\tan{3x}}\)

Jawab:

\(\begin{flalign*} \lim\limits_{x \to 0} \frac{\sin{x}+\tan{3x}-\sin{5x}}{x-\sin{2x}-\tan{3x}} &= \lim\limits_{x \to 0} \frac{\sin{x}+\tan{3x}-\sin{5x}}{x-\sin{2x}-\tan{3x}} \times \frac{{\frac{1}{x}}}{{\frac{1}{x}}} \\ &= \lim\limits_{x \to 0} \frac{\frac{\sin{x}}{x} + \frac{\tan{3x}}{x} - \frac{\sin{5x}}{x}}{1 - \frac{\sin{2x}}{x} - \frac{\tan{3x}}{x}} \\ &= \frac{1+3-5}{1-2-3}=\frac{-1}{-4}=\frac{1}{4} \end{flalign*}\)

5d. Tentukan nilai \(\lim\limits_{x \to 0} \dfrac{\sin{3x}}{2-\sqrt{3x+4}}\)

Jawab:

\(\begin{flalign*} \lim\limits_{x \to 0} \frac{\sin{3x}}{2-\sqrt{3x+4}} &= \lim\limits_{x \to 0} \frac{\sin{3x}}{2-\sqrt{3x+4}} \times \frac{2+\sqrt{3x+4}}{2+\sqrt{3x+4}} \\ &= \lim\limits_{x \to 0} \frac{\sin{3x}\,\left(2+\sqrt{3x+4}\right)}{4-(3x+4)} \\ &= \lim\limits_{x \to 0} \frac{\sin{3x}}{-3x}\times\left(2+\sqrt{3x+4}\right) \\ &= \frac{3}{-2}\times(2+\sqrt{(0+4)})=-\frac{3}{2}\times{4}=-6 \end{flalign*}\)

\(x\to{c}\,\,\,\equiv\,\,\,(x-c)\to{0}\)

Contoh 6

6a. Tentukan nilai \(\lim\limits_{x \to 2} \dfrac{3x\,\sin(2x-4)}{\tan(x-2)}\)

Jawab:

Ingat \(\quad x\to{2}\,\Leftrightarrow\,(x-2)\to{0}\)

Misalkan \(\quad y=x-2\,\Leftrightarrow\, x=y+2\).

Diperoleh:

\(\begin{flalign*} \lim\limits_{x \to 2} \frac{3x\,\sin(2x-4)}{\tan(x-2)} &= \lim\limits_{x \to 2} 3x \times \lim\limits_{(x-2)\to 0} \frac{\sin{2(x-2)}}{\tan(x-2)} \\ &= \lim\limits_{x\to{2}} 3x\times\lim\limits_{y \to 0} \frac{\sin{2y}}{\tan{y}} \\ &= 3\cdot(2)\times\frac{2}{1}=12\end{flalign*}\)

6b. Tentukan nilai \(\lim\limits_{x \to 2} \dfrac{(x^2-7x+10)\,\sin{(x-2)}}{(x^2+x-6)^2}\)

Jawab:

Ingat \(\quad x\to{2}\,\Leftrightarrow\,(x-2)\to{0}\)

\(\begin{flalign*} \lim\limits_{x \to 2} \frac{(x^2-7x+10)\,\sin{(x-2)}}{(x^2+x-6)^2} &= \lim\limits_{x \to 2} \frac{(x-2)\,(x-5)\,\sin{(x-2)}}{((x+3)(x-2))^2} \\ &= \lim\limits_{x \to 2} \frac{(x-2)\,(x-5)\,\sin{(x-2)}}{(x+3)\,(x+3)\,(x-2)\,(x-2)} \\ &= \lim\limits_{x \to 2}\frac{x-5}{(x+3)(x+3)}\times\lim\limits_{(x-2) \to 0}\frac{\sin{(x-2)}}{x-2} \\ &= \frac{2-5}{(2+3)(2+3)}\times{1} = -\frac{3}{25} \end{flalign*}\)

Manipulasi Ekspresi Limit Fungsi

Ingat

  • \(\cos{x}=\sin{(\frac{1}{2}\pi-x)}=\sin{-(x-\frac{1}{2}\pi)}=-\sin{(x-\frac{1}{2}\pi)}\)
  • \(1-\cos^2{ax}=\sin^2{ax}\)
  • \(1-\cos{2ax}=2\sin^2{ax}\)
  • \(\sin{ax}+\sin{bx}=2\,\sin{\frac{1}{2}(a+b)x}\,\cos{\frac{1}{2}(a-b)x}\)
  • \(\sin{ax}-\sin{bx}=2\,\cos{\frac{1}{2}(a+b)x}\,\sin{\frac{1}{2}(a-b)x}\)
  • \(\cos{ax}+\cos{bx}=2\,\cos{\frac{1}{2}(a+b)x}\,\cos{\frac{1}{2}(a-b)x}\)
  • \(\cos{ax}-\cos{bx}=-2\,\sin{\frac{1}{2}(a+b)x}\,\sin{\frac{1}{2}(a-b)x}\)
  • \(\tan{ax}+\tan{bx}=\left(\tan{(ax+bx)}\right)\times\left( 1 - \tan{ax}\,\tan{bx} \right)\)
  • \(\tan{ax}-\tan{bx}=\left(\tan{(ax-bx)}\right)\times\left( 1 + \tan{ax}\,\tan{bx} \right)\)

Contoh 7

7a. Tentukan nilai \(\lim\limits_{x \to \frac{1}{2}\pi} \dfrac{\cos{x}}{x-\frac{1}{2}\pi}\)

\(\begin{flalign*} \cos{x} &= \sin{(\frac{1}{2}\pi-x)} \\ &= \sin{-(x-\frac{1}{2}\pi)} \\ &= -\sin{(x-\frac{1}{2}\pi)} \end{flalign*}\)

dan \(x\to\frac{1}{2}\pi\,\equiv\,(x-\frac{1}{2}\pi)\to{0}\), sehingga

\(\begin{flalign*} \lim\limits_{x\to{\frac{1}{2}\pi}}\frac{\cos{x}}{x-\frac{1}{2}\pi} &= \lim\limits_{(x-\frac{1}{2}\pi)\to{0}}\frac{-\sin{(x-\frac{1}{2}\pi)}}{(x-\frac{1}{2}\pi)} \\ &= -\lim\limits_{(x-\frac{1}{2}\pi)\to{0}}\frac{\sin{(x-\frac{1}{2}\pi)}}{(x-\frac{1}{2}\pi)} \\ &= -1 \end{flalign*}\)

Cara lain

Jika dimisalkan \(\quad y = x - \frac{1}{2}\pi\), maka:

\((x-\frac{1}{2}\pi)\to{0}\) dapat dinyatakan \(y\to{0}\), sehingga

\(\begin{flalign*} \lim\limits_{x \to \frac{1}{2}\pi} \frac{\cos{x}}{x-\frac{1}{2}\pi} &= \lim\limits_{(x-\frac{1}{2}\pi)\to{0}} \frac{\cos{x}}{x-\frac{1}{2}\pi} \\ &= \lim\limits_{y\to{0}} \frac{-\sin{y}}{y} \\ &= -\lim\limits_{y\to{0}} \frac{\sin{y}}{y} \\ &= -1 \end{flalign*}\)

7b. Tentukan nilai \(\lim\limits_{x\to{0}}\dfrac{1-\cos^2{6x}}{4x\,\tan{3x}}\)

\(\begin{flalign*} \lim\limits_{x\to{0}}\frac{1-\cos^2{6x}}{4x\,\tan{3x}} &= \lim\limits_{x\to{0}}\frac{\sin^2{6x}}{4x\,\tan{3x}} \\ &= \lim\limits_{x\to{0}}\frac{\sin{6x}}{4x}\times\lim\limits_{x\to{0}}\frac{\sin{6x}}{\tan{3x}} \\ &= \frac{6}{4}\times\frac{6}{3} \\ &= \frac{3}{2}\times{2} = 3 \end{flalign*}\)

7c. Tentukan nilai \(\lim\limits_{x\to{0}}\dfrac{1-\cos{6x}}{\sin{2x}\,\tan{3x}}\)

\(\begin{flalign*} \lim\limits_{x\to{0}}\frac{1-\cos{6x}}{\sin{2x}\,\tan{3x}} &= \lim\limits_{x\to{0}}\frac{1-\cos{2(3x)}}{\sin{2x}\,\tan{3x}} \\ &= \lim\limits_{x\to{0}}\frac{1-(1-2\sin^2{3x})}{\sin{2x}\,\tan{3x}} \\ &= \lim\limits_{x\to{0}}\frac{2\sin^2{3x}}{\sin{2x}\,\tan{3x}} \\ &= 2\times\lim\limits_{x\to{0}}\frac{\sin{3x}}{\sin{2x}}\times\frac{\sin{3x}}{\tan{3x}} \\ &= 2\times\frac{3}{2}\times{1} = 3 \end{flalign*}\)

7d. Tentukan nilai \(\lim\limits_{x \to {0}}\dfrac{\sin(6x+\frac{1}{3}\pi)+\sin(4x-\frac{1}{3}\pi)}{2x}\)

\(\begin{flalign*} &\lim\limits_{x \to {0}}\frac{\sin(6x+\frac{1}{3}\pi)+\sin(4x-\frac{1}{3}\pi)}{2x} \\ &= \lim\limits_{x \to {0}}\frac{2\sin\frac{1}{2}(6x+\frac{1}{3}\pi+4x-\frac{1}{3}\pi)\,\cos\frac{1}{2}(6x+\frac{1}{3}\pi-(4x-\frac{1}{3}\pi))}{2x} \\ &= \lim\limits_{x \to {0}}\frac{2\sin{5x}\,\cos{(x+\frac{1}{3}\pi)}}{2x} \\ &= \lim\limits_{x \to {0}}\frac{\sin{5x}}{x}\times\cos{(x+\frac{1}{3}\pi)} \\ &= 5\times\cos{\left(0+\frac{1}{3}\pi\right)} = 5\times\cos{\frac{1}{3}\pi} = 5\times\left(\frac{1}{2}\right) = \frac{5}{2} \end{flalign*}\)

7e. Tentukan nilai \(\lim\limits_{\Delta{x} \to {0}}\dfrac{\sin{(a(x + \Delta{x})+b)}-\sin{(ax+b)}}{\Delta{x}}\)

\(\begin{flalign*} &\lim\limits_{\Delta{x} \to {0}}\frac{\sin{(a(x + \Delta{x})+b)}-\sin{(ax+b)}}{\Delta{x}} \\ &= {\small 2\,\lim\limits_{\Delta{x} \to {0}}\frac{\cos\frac{1}{2}(a(x+\Delta{x})+b+ax+b)\,\sin\frac{1}{2}(a(x+\Delta{x})+b-(ax+b))}{\Delta{x}}} \\ &= 2\,\lim\limits_{\Delta{x} \to {0}}\frac{\cos\frac{1}{2}(2ax+a\Delta{x}+2b)\,\sin\frac{1}{2}a\Delta{x}}{\Delta{x}} \\ &= 2\,\lim\limits_{\Delta{x} \to {0}}{\cos\left(ax+b+\frac{1}{2}a\Delta{x}\right)}\times \lim\limits_{\Delta{x} \to {0}}\frac{\sin\frac{1}{2}a\Delta{x}}{\Delta{x}} \\ &= 2\,\cos{(ax+b+0)}\times\frac{1}{2}a \\ &= a\,\cos{(ax+b)} \end{flalign*}\)

7f. Tentukan nilai \(\lim\limits_{x \to {0}}\dfrac{\cos(4x+\pi)+\cos{2x}}{x^2}\)

\(\begin{flalign*} &\lim\limits_{x \to {0}}\frac{\cos(4x+\pi)+\cos{2x}}{x^2} \\ &= \lim\limits_{x \to {0}}\frac{2\,\cos\frac{1}{2}(4x+\pi+2x)\,\cos\frac{1}{2}(4x+\pi-2x)}{x^2} \\ &= 2\,\lim\limits_{x \to {0}}\frac{\cos{(\frac{1}{2}\pi+3x)}}{x}\times\frac{\cos{(\frac{1}{2}\pi+x)}}{x} \\ &= 2\,\lim\limits_{x \to {0}}\frac{-\sin{3x}}{x}\times\frac{-\sin{x}}{x} \\ &= 2\,\times{(-3)}\times{(-1)} = 6 \end{flalign*}\)

7g. Tentukan nilai \(\lim\limits_{\Delta{x} \to {0}}\dfrac{\cos{(a(x + \Delta{x})+b)}-\cos{(ax+b)}}{\Delta{x}}\)

\(\begin{flalign*} &\lim\limits_{\Delta{x} \to {0}}\frac{\cos{(a(x + \Delta{x})+b)}-\cos{(ax+b)}}{\Delta{x}} \\ &= {\small \lim\limits_{\Delta{x} \to {0}}\frac{-2\,\sin\frac{1}{2}(a(x+\Delta{x})+b+ax+b)\,\sin\frac{1}{2}(a(x+\Delta{x})+b-(ax+b))}{\Delta{x}}} \\ &= -2\,\lim\limits_{\Delta{x} \to {0}}\frac{\sin\frac{1}{2}(2ax+a\Delta{x}+2b)\,\sin\frac{1}{2}a\Delta{x}}{\Delta{x}} \\ &= -2\,\lim\limits_{\Delta{x} \to {0}}{\sin\left(ax+b+\frac{1}{2}a\Delta{x}\right)}\times \lim\limits_{\Delta{x} \to {0}}\frac{\sin\frac{1}{2}a\Delta{x}}{\Delta{x}} \\ &= -2\,\sin{(ax+b+0)}\times\frac{1}{2}a \\ &= -a\,\sin{(ax+b)} \end{flalign*}\)

7h. Tentukan nilai \(\lim\limits_{x \to {0}}\dfrac{\tan(5x+\frac{1}{6}\pi)+\tan(3x-\frac{1}{6}\pi)}{2x}\)

\(\begin{flalign*} &\lim\limits_{x \to {0}}\frac{\tan(5x+\frac{1}{6}\pi)+\tan(3x-\frac{1}{6}\pi)}{2x} \\ &= \lim\limits_{x \to {0}}\frac{\tan{(5x+\frac{1}{6}\pi+3x+\frac{1}{6}\pi)}\times(1-\tan{(5x+\frac{1}{6}\pi)}\,\tan{(3x+\frac{1}{6}\pi)})}{2x} \\ &= \lim\limits_{x \to {0}}\frac{\tan{(8x+\frac{1}{3}\pi)}}{2x}\times\lim\limits_{x \to {0}}\left(1-\tan{\left(5x+\frac{1}{6}\pi\right)}\,\tan{\left(3x+\frac{1}{6}\pi\right)}\right) \\ &= \tan{\frac{1}{3}\pi}\times\left(1-\left(\tan{\frac{1}{6}\pi}\,\tan{\frac{1}{6}\pi}\right)\right) \\ &= \sqrt{3}\times\left(1-\left(\frac{1}{3}{\sqrt{3}}\times\frac{1}{3}{\sqrt{3}}\right)\right) \\ &= \sqrt{3}\times\left(1-\frac{1}{3}\right) = \frac{2}{3}\sqrt{3} \end{flalign*}\)

7i. Tentukan nilai \(\lim\limits_{\Delta{x} \to {0}}\dfrac{\tan{(a(x + \Delta{x})+b)}-\tan{(ax+b)}}{\Delta{x}}\)

\(\begin{flalign*} &\lim\limits_{\Delta{x} \to {0}}\frac{\tan{(a(x + \Delta{x})+b)}-\tan{(ax+b)}}{\Delta{x}} \\ &= {\small \lim\limits_{\Delta{x} \to {0}}\frac{\tan{(a(x + \Delta{x})+b-(ax+b))}\times(1+\tan{(a(x + \Delta{x})+b)}\,\tan{(ax+b)})}{\Delta{x}}} \\ &= \lim\limits_{\Delta{x} \to {0}}\frac{\tan{a\Delta{x}}}{\Delta{x}}\times\lim\limits_{\Delta{x} \to {0}}\left(1+\tan{(a(x + \Delta{x})+b)}\,\tan{(ax+b)}\right) \\ &= a\times\left(1+\tan{(ax+b)}\,\tan{(ax+b)}\right) = a\times\left(1+\tan^2{(ax+b)}\right) \\ &= a\times\left(\frac{\cos^2{(ax+b)}}{\cos^2{(ax+b)}}+\frac{\sin^2{(ax+b)}}{\cos^2{(ax+b)}}\right) \\ &= a\times\frac{\cos^2{(ax+b)}+\sin^2{(ax+b)}}{\cos^2{(ax+b)}} \\ &= a\times\frac{1}{\cos^2{(ax+b)}} \\ &= a\,\sec^2{(ax+b)} \end{flalign*}\)

Soal Jawab Limit Fungsi Trigonometri

\(\text{01a. } \lim\limits_{x\to\frac{1}{4}\pi} \dfrac{2-\cos{2x}}{\tan{x}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to\frac{1}{4}\pi} \frac{2-\cos{2x}}{\tan{x}} &= \frac{2-\cos{(2\cdot\frac{1}{4}\pi)}}{\tan{\frac{1}{4}\pi}} \\ &= \frac{2-\cos{(\frac{1}{2}\pi)}}{\tan{\frac{1}{4}\pi}} \\ &= \frac{2-0}{1} = 2 \\ \end{flalign*}\)

\(\text{01b. } \lim\limits_{x\to\frac{1}{3}\pi} \dfrac{\sin{x}\times\cos{x}}{\sin^2{x}-\cos^2{x}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to\frac{1}{3}\pi} \frac{\sin{x}\times\cos{x}}{\sin^2{x}-\cos^2{x}} &= \frac{\sin{\frac{1}{3}\pi}\times\cos{\frac{1}{3}\pi}}{\sin^2{\frac{1}{3}\pi}-\cos^2{\frac{1}{3}\pi}} \\ &= \frac{\frac{1}{2}\sqrt{3}\times\frac{1}{2}}{\left(\frac{1}{2}\sqrt{3}\right)^2-\left(\frac{1}{2}\right)^2} \\ &= \frac{\frac{1}{4}\sqrt{3}}{\frac{3}{4}-\frac{1}{4}} \\ &= \frac{\frac{1}{4}\sqrt{3}}{\frac{2}{4}} \\ &= \frac{1}{2}\sqrt{3} \end{flalign*}\)

\(\text{02a. } \lim\limits_{x\to{0}} \dfrac{\tan{2x}}{2x-\sin{x}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{0}} \frac{\tan{2x}}{2x-\sin{x}} &= \lim\limits_{x\to{0}} \frac{\tan{2x}}{2x-\sin{x}}\times\frac{\frac{1}{x}}{\frac{1}{x}} \\ &= \lim\limits_{x\to{0}} \frac{\frac{\tan{2x}}{x}}{2-\frac{\sin{x}}{x}} \\ &= \frac{2}{2-1} = 2 \end{flalign*}\)

\(\text{02b. } \lim\limits_{x\to{0}} \dfrac{5x+\tan{3x}}{3x-\sin{x}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{0}} \frac{5x+\tan{3x}}{3x-\sin{x}} &= \lim\limits_{x\to{0}} \frac{5x+\tan{3x}}{3x-\sin{x}}\times\frac{\frac{1}{x}}{\frac{1}{x}} \\ &= \lim\limits_{x\to{0}} \frac{5+\frac{\tan{3x}}{x}}{3-\frac{\sin{x}}{x}} \\ &= \frac{5+3}{3-1} = 4 \end{flalign*}\)

\(\text{03a. } \lim\limits_{x\to{0}} \dfrac{4x\,\cos{2x}+2\tan{3x}}{3x-\sin{x}\cos{x}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{0}} \frac{4x\,\cos{2x}+2\tan{3x}}{3x-\sin{x}\cos{x}} &= \lim\limits_{x\to{0}} \frac{4x\,\cos{2x}+2\tan{3x}}{3x-\sin{x}\cos{x}}\times\frac{\frac{1}{x}}{\frac{1}{x}} \\ &= \lim\limits_{x\to{0}} \frac{4\,\cos{2x}+2\times\frac{\tan{3x}}{x}}{3-\frac{\sin{x}}{x}\times\cos{x}} \\ &= \frac{4\,\cos{0} + 2\times{3}}{3-1\times\cos{0}} \\ &= \frac{4\times{1}+6}{3-1\times{1}} \\ &= \frac{10}{2}=5 \end{flalign*}\)

\(\text{03b. } \lim\limits_{x\to{0}} \dfrac{\cos{4x}\,\sin{3x}+2\tan{3x}}{5x-2\sin{x}\cos{x}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{0}} \frac{\cos{4x}\,\sin{3x}+2\tan{3x}}{5x-2\sin{x}\cos{x}} &= \lim\limits_{x\to{0}} \frac{\cos{4x}\,\sin{3x}+2\tan{3x}}{5x-2\sin{x}\cos{x}}\times\frac{\frac{1}{x}}{\frac{1}{x}} \\ &= \lim\limits_{x\to{0}} \frac{\cos{4x}\times\frac{\sin{3x}}{x}+2\times\frac{\tan{3x}}{x}}{5-2\times\frac{\sin{x}}{x}\times\cos{x}} \\ &= \frac{\cos{0}\times{3} + 2\times{3}}{5-2\times{1}\times\cos{0}} \\ &= \frac{1\times{3}+6}{5-2\times{1}\times{1}} \\ &= \frac{9}{3}=3 \end{flalign*}\)

\(\text{04a. } \lim\limits_{x\to{3}} \dfrac{\sin(6x-18)}{3x^2 - 9x} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{3}} \frac{\sin(6x-18)}{3x^2 - 9x} &= \lim\limits_{x\to{3}} \frac{\sin{6(x-3)}}{3x(x-3)} \\ &= \lim\limits_{x\to{3}} \frac{1}{3x} \times \lim\limits_{x\to{3}} \frac{\sin{6(x-3)}}{(x-3)} \\ &= \frac{1}{3\times{3}} \times 6 \\ &= \frac{6}{3\times{3}} = \frac{2}{3} \end{flalign*}\)

\(\text{04b. } \lim\limits_{x\to{-1}} \dfrac{\sin(4x+4)}{2x^2 + 2x} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{-1}} \frac{\sin(4x+4)}{2x^2+2x} &= \lim\limits_{x\to{-1}} \frac{\sin{4(x+1)}}{2x(x+1)} \\ &= \lim\limits_{x\to{-1}} \frac{1}{2x} \times \lim\limits_{x\to{-1}} \frac{\sin{4(x+1)}}{(x+1)} \\ &= \frac{1}{2\times{(-1)}} \times 4 \\ &= \frac{4}{-2} = -2 \end{flalign*}\)

\(\text{05a. } \lim\limits_{x\to{2}} \dfrac{1-\cos(6x-12)}{3x^2-12x+12} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{2}} \frac{1-\cos(6x-12)}{3x^2-12x+12} &= \lim\limits_{x\to{2}} \frac{1-\cos{6(x-2)}}{3{(x^2-4x+4)}} \\ &= \lim\limits_{x\to{2}} \frac{2\sin^2{3(x-2)}}{3{(x-2)^2}} \\ &= \frac{2}{3}\times\lim\limits_{x\to{2}} \frac{\sin{3(x-2)}}{{(x-2)}} \times \lim\limits_{x\to{2}} \frac{\sin{3(x-2)}}{{(x-2)}} \\ &= \frac{2}{3}\times{3}\times{3} \\ &= 6 \end{flalign*}\)

\(\text{05b. } \lim\limits_{x\to{2}} \dfrac{1-\cos^2(6x-12)}{3x^2-12x+12} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{2}} \frac{1-\cos^2(6x-12)}{3x^2-12x+12} &= \lim\limits_{x\to{2}} \frac{1-\cos^2{6(x-2)}}{3{(x^2-4x+4)}} \\ &= \lim\limits_{x\to{2}} \frac{\sin^2{6(x-2)}}{3{(x-2)^2}} \\ &= \frac{1}{3}\times\lim\limits_{x\to{2}} \frac{\sin{6(x-2)}}{{(x-2)}} \times \lim\limits_{x\to{2}} \frac{\sin{6(x-2)}}{{(x-2)}} \\ &= \frac{1}{3}\times{6}\times{6} \\ &= 12 \end{flalign*}\)

Soal Pengayaan dan Solusinya

\(\text{01. } \lim\limits_{x\to{0}} \dfrac{x^3}{\sin{2x}-\tan{2x}} = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{0}} \frac{x^3}{\sin{2x}-\tan{2x}} &= \lim\limits_{x\to{0}} \frac{x^3}{\sin{2x}-\frac{\sin{2x}}{\cos{2x}}} \\ &= \lim\limits_{x\to{0}} \frac{x^3}{\sin{2x}\left(1-\frac{1}{\cos{2x}}\right)} \\ &= \lim\limits_{x\to{0}} \frac{x^3}{\sin{2x}\left(\frac{\cos{2x}-1}{\cos{2x}}\right)} \\ &= \lim\limits_{x\to{0}} \frac{x^3}{\sin{2x}({\cos{2x}-1})} \\ &= \lim\limits_{x\to{0}} \frac{x^3\,\cos{2x}}{\sin{2x}\left(1-2\sin^2{x}-1\right)} \\ &= \lim\limits_{x\to{0}} \frac{x^3\,\cos{2x}}{\sin{2x}(-2\sin^2{x})}\\ &= -\frac{1}{2}\times\lim\limits_{x\to{0}} \frac{x}{\sin{2x}}\times\frac{x}{\sin{x}}\times\frac{x}{\sin{x}}\times\cos{2x} \\ &= -\frac{1}{2}\times\frac{1}{2}\times{1}\times{1}\times\cos{0} \\ &= -\frac{1}{4} \end{flalign*}\)

\(\text{02. } \lim\limits_{x\to{\frac{1}{2}\pi}} (\sec^2{x} - \sec{x}\cdot\tan{x}) = \cdots.\)

\(\begin{flalign*} \lim\limits_{x\to{\frac{1}{2}\pi}} (\sec^2{x} - \sec{x}\cdot\tan{x}) &= \lim\limits_{x\to{\frac{1}{2}\pi}} \left(\frac{1}{\cos^2{x}} - \frac{1}{\cos{x}}\cdot\frac{\sin{x}}{\cos{x}}\right) \\ &= \lim\limits_{x\to{\frac{1}{2}\pi}} \left(\frac{1-\sin{x}}{\cos^2{x}}\right) \\ &= \lim\limits_{x\to{\frac{1}{2}\pi}} \left(\frac{1-\sin{x}}{1-\sin^2{x}}\right) \\ &= \lim\limits_{x\to{\frac{1}{2}\pi}} \left(\frac{1-\sin{x}}{(1+\sin{x})(1-\sin{x})}\right) \\ &= \lim\limits_{x\to{\frac{1}{2}\pi}} \left(\frac{1}{1+\sin{x}}\right) \\ &= \frac{1}{1+\sin{\frac{1}{2}\pi}} \\ &= \frac{1}{1+1} = \frac{1}{2} \\ \end{flalign*}\)

\(\text{03. } \text{Tentukan nilai }\lim\limits_{x\to{0}} \left[ \csc^2(2x) - \dfrac{1}{4x^2} \right]\)

\[ \begin{aligned} &\lim_{x\to{0}} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\ &= \lim_{x\to{0}} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\ &= \lim_{x\to{0}} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] &{\small\sin^2(2x) = 4x^2\left(\frac{\sin^2(2x)}{2x}\right)^2}\\ &= \lim_{x\to{0}} \left[ \frac{4x^2-\sin^2(2x)}{4x^2\times 4x^2\left(\frac{\sin(2x)}{2x}\right)^2} \right] \\ &= \lim_{x\to{0}} \left[ \frac{4x^2-\sin^2(2x)}{4x^2\times 4x^2} \right] \times \lim_{x\to{0}} \left[ \frac{1}{\left(\frac{\sin(2x)}{2x}\right)^2} \right] \\ &= \lim_{x\to{0}} \left[ \frac{4x^2-\sin^2(2x)}{16x^4} \right] \times 1 \\ &= \lim_{x\to{0}} \left[ \frac{8x-2\sin(4x)}{64x^3} \right] \\ &= \lim_{x\to{0}} \left[ \frac{8-8\cos(4x)}{192x^2} \right] \\ &= \lim_{x\to{0}} \left[ \frac{1-\cos(4x)}{24x^2} \right] \\ &= \frac{1}{24} \times \lim_{x\to{0}} \left[ \frac{2\sin^2{2x}}{x^2} \right] \\ &= \frac{1}{24} \times 2 \times 2^2 \\ &= \frac{1}{3} \\ \end{aligned} \]

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