Question

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

• A. $e^2$
• B. e
• C. 1
• D. 1/e

Answer 1

Answer: (C)

1

$For{lim}_{x\to \infty }x[\log x+1-\log x]={lim}_{x\to \infty }x\left[\log \frac{x+1}{x}\right]={lim}_{x\to \infty }\frac{\left[\log (1+\frac{1}{x})\right]}{\frac{1}{x}}\Rightarrow Asx\to \infty ,1/x\to 0Therefore,{lim}_{x\to \infty }\frac{\left[\log (1+\frac{1}{x})\right]}{\frac{1}{x}}={lim}_{x\to 0}\frac{\left[\log (1+x)\right]}{x}\left(Replacing1/xbyx\right)Now,{lim}_{x\to 0}\frac{\left[\log (1+x)\right]}{x}=\frac{0}{0}Nowusing{L}^{\prime }hospita{l}^{\prime }srule,\Rightarrow {lim}_{x\to 0}\frac{\left[\log (1+x)\right]}{x}={lim}_{x\to 0}\frac{d\left(\left[\log (1+x)\right]\right)/dx}{dx/dx}={lim}_{x\to 0}\frac{\frac{1}{1+x}}{1}=\frac{1}{1}=1$