$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x + \log (1 - x)}}{{{x^2}}}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to \pi /2} \tan x \log \sin x = $

$\mathop {\lim }\limits_{x \to 0} x \log (\sin x) = $

$\lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = $

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

ધારો કે $f(x) = \lim_{n \rightarrow \infty} \sum_{r=0}^n \left( \frac{2\tan(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})} \right)$. તો $\lim_{x \rightarrow 0} \frac{e^x - e^{f(x)}}{x - f(x)}$ ની કિંમત . . . . . . . છે.