Let $f(x) = \frac{\sqrt{x+3}}{x+1}$. Then the value of $\lim_{x \rightarrow -3^{-}} f(x)$ is

Evaluate: $\mathop {\lim }\limits_{x \to 1} \frac{x^{15}-1}{x^{10}-1}$

$\lim _{n \rightarrow \infty} \frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\ldots+\frac{n^2}{n^3+1} = $

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to-

If $\lim _{n \rightarrow \infty} \frac{1-(10)^n}{1+(10)^{n+1}}=\frac{-\alpha}{10}$,then $\alpha$ is equal to

If $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{a^{1/x}} + b}}{c}} \right)^x} = d$ (where $d$ is a non-zero finite value),then $(b + 1) \log_a d$ is equal to: