$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ is

Evaluate the limit: $\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$

If $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$ and $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$,then $12 k=$

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 :

The value of $\mathop {\lim }\limits_{x \to a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$ is

$\mathop {\lim }\limits_{x \to \pi /6} \left[ {\frac{{3\sin x - \sqrt 3 \cos x}}{{6x - \pi }}} \right] = $