Question

Solve $\underset{x\to \infty }{lim}\frac{x+\log x}{x^2+\sin x}$

• A. $0$
• B. $1$
• C. $\infty$
• D. does not exits

Answer 1

Answer: (A)

$0$

$\underset{x\to \infty }{lim}\frac{x+\log x}{x^2+\sin x}$

Put $x=\frac{1}{x}$ then $x\to \infty$ and $x\to 0$

$\underset{x\to \infty }{lim}\frac{\frac{1}{x}+\log \frac{1}{x}}{\frac{1}{x^2}+\sin \frac{1}{x}}=\underset{x\to \infty }{lim}\frac{[1+(\log \frac{1}{x}\left)x\right]}{1+x^2\sin \frac{1}{x}}$

$=\underset{x\to \infty }{lim}\frac{0\times \left[\log \right(\frac{1}{x})\times 0+1]}{1+0}$

$=0$