# How to show that $\lim_{x\to \infty} \left(\int^x_2 (\ln t)^{-1} dt \right) \big/ (x\ /\ln x)=1$?

Show that $$\lim_{x\to \infty}\frac{\int^x_2\frac{1}{\ln t}dt}{\frac{x}{\ln x}}=1.$$

I thought to use L'Hospital's rule, but for that both denominator and numerator should go to infinity. I am not convinced my self for that.

My attempt:

By applying L'Hospital's rule, we get $$\lim_{x \to \infty} \frac{1/\ln x}{\frac{\ln x-1}{(\ln x)^2}}.$$ Here, for denominator I again use L'Hospital's rule, to get $$\lim_{x \to \infty}\frac{\ln x}{\ln x-1}=1.$$ Am I right? And what are reasons for the numerator to tend to infinity?

Any help will be appreciated.

• There is no need to check whether numerator goes to infinity. Denominator tending to infinity is sufficient for the application of L'Hospital's Rule. Sep 18, 2018 at 13:15
• @ParamanandSingh Sir Are you mean after log x shift to numerator na? Sep 18, 2018 at 13:36
• I am talking of the original expression (where there is an integral in numerator). Sep 18, 2018 at 13:42
• The case $0/0$ for L'Hospital's Rule is popular, but the other case is $\text{anything} /\infty$ which is not so popular. Sep 18, 2018 at 13:47
• Sir I am Knowing the rule that if both Numerator and demominator both goes to 0 or infity then only we can apply lhopital .Is I am wrong @ParamanandSingh? Sep 18, 2018 at 13:49

$\displaystyle\int^x_2\frac{1}{\ln t}\,\mathrm dt$ tends to $+\infty$ because $\;\ln t<t$ for all $t$, so if $t>1$, $$\frac1{\ln t}>\frac 1t,\;\text{whence }\;\int^x_2\frac{1}{\ln t}\,\mathrm dt\ge \int^x_2\frac{1}{t}\,\mathrm dt=\ln x-\ln 2.,$$

• Thanks A lot Sir. Is my Application of L'Hopital rule is right? Sep 18, 2018 at 12:53
• Yes, absolutely. In my opinion, the second application of L'Hospital is unnecessary. Sep 18, 2018 at 12:54