If $f(x)=\left\{\begin{array}{ll}x+2, & x \leq-1 \\ c x^2, & x>-1\end{array}\right.$, then find $c$ when $\lim _\limits{x \rightarrow-1} f(x)$ exists.
$$\begin{aligned} \text{Given,}\quad f(x) & = \begin{cases}x+2, & x \leq-1 \\ c x^2, & x>-1\end{cases} \\ \mathrm{LHL} & =\lim _{x \rightarrow--^{-}} f(x)=\lim _{x \rightarrow-1^{-}}(x+2) \\ & =\lim _{h \rightarrow 0}(-1-h+2)=\lim _{h \rightarrow 0}(1-h)=1 \\ \mathrm{RHL} & =\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}} c x^2=\lim _{h \rightarrow 0} c(-1+h)^2 \\ \therefore\quad & =c \end{aligned}$$
$$\begin{aligned} &\text { If } \lim _{x \rightarrow-1} f(x) \text { exist, then } \mathrm{LHL}=\mathrm{RHL}\\ &\therefore \quad c=1 \end{aligned}$$
$\lim _\limits{x \rightarrow \pi} \frac{\sin x}{x-\pi}$ is equal to
$\lim _\limits{x \rightarrow 0} \frac{x^2 \cos x}{1-\cos x}$ is equal to
$\lim _\limits{x \rightarrow 0} \frac{(1+x)^n-1}{x}$ is equal to
$\lim _\limits{x \rightarrow 1} \frac{x^m-1}{x^n-1}$ is equal to