Question

Let $l=\underset{ x \rightarrow \infty}{\text{Lim}} \int\limits_{ x }^{2 x } \frac{ dt }{ t }$ and $m =\underset{ x \rightarrow \infty}{\text{Lim}} \frac{1}{ x \ln x } \int\limits_1^{ x } \ln tdt$ then the correct statement is

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

Answer 1

Answer: (A)

$l=\underset{ x \rightarrow \infty}{\text{Lim}} \ln 2 x -\ln x =\ln 2 ; m =\frac{\int\limits_1^{ x } \ln t }{ x \ln x }=\underset{ x \rightarrow \infty}{\text{Lim}}\frac{\ln x }{ x \cdot \frac{1}{ x }+\ln x }=\underset{ x \rightarrow \infty}{\text{Lim}}\frac{\ln x }{1+\ln x }=1$
Hence $l \times m =\ln 2 \cdot 1=\ln 2=l$