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

$\lim _{x \rightarrow 1} \frac{\tan \left(x^{2}-1\right)}{x-1}$ is equal to

If $f$ is a strictly increasing function,then $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$ is equal to

If $\mathop {\lim }\limits_{x \to 0} \frac{{\log (3 + x) - \log (3 - x)}}{x} = k,$ then the value of $k$ is

Let $f(1) = g(1) = k$ and their $n^{th}$ derivatives $f^{(n)}(1), g^{(n)}(1)$ exist and are not equal for some $n$. If $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$,then the value of $k$ is:

The value of $\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$ is