$\mathop {\lim }\limits_{x \to 1} \frac{{\log x}}{{x - 1}} = $

Evaluate $\lim _{x \rightarrow 0^{+}} (x^{n} \ln x)$ for $n > 0$.

The value of $\mathop {\lim }\limits_{x \to \pi /4} \frac{{{{\cot }^3}x - \tan x}}{{\cos \left( {x + \pi /4} \right)}}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \sin x}}{{{x^2}\sin x}} = $

If $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$,$l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \pi/2} \frac{\cos x}{x - \pi/2}$,then:

Let $f: R \rightarrow R$ be a continuous function. Then $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(t) dt}{x^{2}-\frac{\pi^{2}}{16}}$ is equal to :