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