Berikut adalah contoh soal dan langkah penyelesaiannya:
Contoh 1
\(\color{purple}\lim\limits_{x \rightarrow \infty} \dfrac{2x^2 \tan \left(\dfrac{1}{x}\right)\:-\:x \sin \left(\dfrac{1}{x}\right) + \dfrac{1}{x}}{x\cos \left(\dfrac{2}{x}\right)} = \dotso\)
Solusi:
Misalkan \(\dfrac{1}{x} = y\)
Jika x mendekati nilai takhingga maka y akan mendekati nilai nol. Jadi kita dapat mengubah soalnya menjadi limit y mendekati nol.
Karena \(\dfrac{1}{x} = y\), maka kita ubah semua variabel x menjadi variabel y. \(x = \dfrac{1}{y}\)
\(\lim\limits_{y \rightarrow 0} \dfrac{2\left(\dfrac{1}{y}\right)^2 \tan y\:-\:\dfrac{1}{y} \sin y + y}{\dfrac{1}{y}\cos 2y}\)
\(\lim\limits_{y \rightarrow 0} \dfrac{2\cdot\dfrac{1}{y}\cdot \dfrac{\tan y}{y}\:-\:\dfrac{\sin y}{y} + y}{\dfrac{1}{y}\cos 2y}\)
Ingat:
\(\color{blue} \lim\limits_{y \rightarrow 0} \dfrac{\tan y}{y} = 1\)
\(\color{blue} \lim\limits_{y \rightarrow 0} \dfrac{\sin y}{y} = 1\)
\(\color{blue} \cos 0 = 1\)
\(\lim\limits_{y \rightarrow 0} \dfrac{2\cdot\dfrac{1}{y}\cdot 1\:-\:1 + y}{\dfrac{1}{y}\cdot 1}\)
Kalikan pembilang dan penyebut dengan y,
\(\lim\limits_{y \rightarrow 0} \:(2\:-\:y + y^2)\)
\(2\:-\:0 + 0^2\)
\(2\)
Jadi,
\(\lim\limits_{x \rightarrow \infty} \dfrac{2x^2 \tan \left(\dfrac{1}{x}\right)\:-\:x \sin \left(\dfrac{1}{x}\right) + \dfrac{1}{x}}{x\cos \left(\dfrac{2}{x}\right)} = 2\)
Contoh 2
\(\color{purple}\lim\limits_{x \rightarrow \infty} \dfrac{\left(1\:-\:\cos \dfrac{2}{x}\right)\cdot x^2 \cdot \sin \dfrac{1}{x}}{\sin \dfrac{4}{x}} = \dotso\)
Solusi:
Misalkan \(\dfrac{1}{x} = y\)
Jika x mendekati nilai takhingga maka y akan mendekati nilai nol. Jadi kita dapat mengubah soalnya menjadi limit y mendekati nol.
Karena \(\dfrac{1}{x} = y\), maka kita ubah semua variabel x menjadi variabel y. \(x = \dfrac{1}{y}\)
\(\lim\limits_{y \rightarrow 0} \dfrac{(1\:-\:\cos 2y)\cdot \left(\dfrac{1}{y}\right)^2 \cdot \sin y}{\sin 4y}\)
Ingat:
\(\color{blue} \cos 2y = 1\:-\:2\sin^2 y\)
\(\color{blue} \lim\limits_{y \rightarrow 0} \dfrac{\sin^2 y}{y^2} = 1\)
\(\color{blue} \lim\limits_{y \rightarrow 0} \dfrac{\sin ay}{\sin by} = \dfrac{a}{b}\)
\(\lim\limits_{y \rightarrow 0} \dfrac{[1\:-\:(1\:-\:2\sin^2 y)]\cdot \left(\dfrac{1}{y}\right)^2 \cdot \sin y}{\sin 4y}\)
\(\lim\limits_{y \rightarrow 0} \dfrac{2\cdot\sin^2 y\cdot \left(\dfrac{1}{y}\right)^2 \cdot \sin y}{\sin 4y}\)
\(\lim\limits_{y \rightarrow 0} \dfrac{2\cdot \dfrac{\sin^2 y}{y^2}\cdot \sin y}{\sin 4y}\)
\(\lim\limits_{y \rightarrow 0} \dfrac{2\cdot 1^2\cdot \sin y}{\sin 4y}\)
\(2\cdot \lim\limits_{y \rightarrow 0} \dfrac{\sin y}{\sin 4y}\)
\(2\cdot \dfrac{1}{4}\)
\(\dfrac{1}{2}\)
Jadi, \(\lim\limits_{x \rightarrow \infty} \dfrac{\left(1\:-\:\cos \dfrac{2}{x}\right)\cdot x^2 \cdot \sin \dfrac{1}{x}}{\sin \dfrac{4}{x}} = \dfrac{1}{2}\)