Question

Let $l=\underset{x\to \infty }{\mathrm{Lim}}\underset{x}{\overset{2x}{\int }}\frac{dt}{t}$ and $m=\underset{x\to \infty }{\mathrm{Lim}}\frac{1}{x\ln x}\underset{1}{\overset{x}{\int }}\ln tdt$ then the correct statement is

• A. %l m =l$
• B. $lm=m$
• C. $l=m$
• D. $l>m$

Answer 1

Answer: (A)

$l=\underset{x\to \infty }{\mathrm{Lim}}\ln 2x-\ln x=\ln 2;m=\frac{\underset{1}{\overset{x}{\int }}\ln t}{x\ln x}=\underset{x\to \infty }{\mathrm{Lim}}\frac{\ln x}{x\cdot \frac{1}{x}+\ln x}=\underset{x\to \infty }{\mathrm{Lim}}\frac{\ln x}{1+\ln x}=1$
Hence $l\times m=\ln 2\cdot 1=\ln 2=l$