Question

$\underset{x\to \infty }{Lt}x\left[log\right(x+1)-logx]=$

• A. $e^2$
• B. $e$
• C. $1$
• D. $1/e$

Answer 1

The correct option is C $1$

Now,

$\underset{x\to \infty }{Lt}x\left[\log \right(x+1)-\log x]$

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

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

$=\underset{y\to 0}{Lt}\left[\frac{\log (1+y)}{y}\right]$ [ Taking $y=\frac{1}{x}$]

$=1$.