<s>[INST] How to find the limits $\lim\limits_{h\rightarrow 0} \frac{e^{-h}}{-h}$ and $\lim\limits_{h\rightarrow 0} \frac{|\cos h-1|}{h}$? How to work around to find the limit for these functions : 


*

*$$\lim_{h\rightarrow 0} \frac{e^{-h}}{-h}$$

*$$\lim_{h\rightarrow 0} \frac{|\cos h-1|}{h}$$
For the second one i think that the limit doesn't exist. 
 [/INST] We know that $e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$ Note that for $|h|<1$ and sufficiently small, 
$$
\frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1
\leq
e^{-h}
\leq 
\frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1+\frac{1}{2!}(-h)^2
$$
which is the same as
$$
1-h
\leq
e^{-h}
\leq 
1-h+\frac{1}{2}h^2
$$
implies
$$
1-\frac{1}{h}
\geq 
\frac{e^{-h}}{-h}
\geq
1-\frac{1}{h}-\frac{1}{2}h
$$
Make $h\to 0$. What you get? Now the other limit. Note that 
\begin{align}
\frac{|\cos(h)-1|}{h} 
=
&
2\frac{\left|\cos\left(2\dfrac{h}{2}\right)-1\right|}{\dfrac{h}{2}}
\\
=
&
2\frac{\left|1-2\sin^2\left(\dfrac{h}{2}\right)-1\right|}{\left(\dfrac{h}{2}\right)}
\\
=
&
4\frac{\left|\sin^2\left(\dfrac{h}{2}\right)\right|}{\left(\dfrac{h}{2}\right)}
\\
=
&
4\left(\dfrac{h}{2}\right)\frac{\sin^2\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)^2}
\end{align}
Make $h\to 0$. What you get?
</s>