Pembahasan Soal Limit Fungsi 1

Nilai dari $\lim _{x\rightarrow \infty }(x^3\sin \frac{1}{x}+x)(\frac{1}{x-1}-\frac{1}{x+1})=\cdots$

Pembahasan

$\lim _{x\rightarrow \infty }(x^3\sin \frac{1}{x}+x)(\frac{1}{x-1}-\frac{1}{x+1})=L$

Permasalahan diatas harus kita bawa ke bentuk berikut

$\lim_{x\rightarrow 0} \frac{\sin nx}{nx}=\lim_{x \rightarrow 0} \frac{nx}{\sin nx}=1$

atau

$\lim_{x\rightarrow \infty} \frac{\sin \frac{1}{nx}}{\frac{1}{nx}}=\lim_{x \rightarrow \infty} \frac{\frac{1}{nx}}{\sin \frac{1}{nx}}=1$

atau

$\lim_{x\rightarrow \infty} nx \sin \frac{1}{nx}=\lim_{x \rightarrow \infty} \frac{1}{nx\sin \frac{1}{nx}}=1$

Mari Kita proses

misalkan $p=\frac{1}{x}$ dan $x\rightarrow \infty$ maka $p\rightarrow 0$

Dengan demikian soal diatas dapat kita tuliskan menjadi

$L=\lim _{p\rightarrow 0 }(\frac{1}{p^3}\sin p +\frac{1}{p})(\frac{1}{\frac{1}{p}-1}-\frac{1}{\frac{1}{p}+1})$

$L=\lim _{p\rightarrow 0 }(\frac{1}{p^2}\frac{\sin p}{p} +\frac{1}{p})(\frac{p}{1-p}-\frac{p}{1+p})$

$L=\lim _{p\rightarrow 0 }(\frac{1}{p^2}+\frac{1}{p})(\frac{2p^2}{1-p^2})$

$L=\lim _{p\rightarrow 0 }(\frac{1+p}{p^2})(\frac{2p^2}{1-p^2})$

$L=\lim _{p\rightarrow 0 }\frac{2(1+p)}{1-p^2}$

$L=\lim _{p\rightarrow 0 }\frac{2}{1-p}$

$L=2$