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

લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x (\tan^{-1} t)^2 dt}}{{\sin x - x}}$

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = $

$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ ની કિંમત શું છે?

$\mathop {\lim }\limits_{\alpha \to \beta } \left[ {\frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}} \right] = $

જો $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$ અને $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$ હોય,તો $12 k=$