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. 
 A: HINT:
$(1):\lim_{h\to0}e^{-h}=1$
$(2):$ $$\cos h=1-2\sin^2\frac h2\implies \cos h-1=-2\sin^2\frac h2$$
$$\implies \frac{\cos h-1}h=-\left(\frac{\sin \frac h2}{\frac h2}\right)^2 \frac h4$$
A: Note that $\cos h\le 1$ so $|\cos h-1|=1-\cos h$.
$$\lim_{h\to 0} \frac{1-\cos h}{h}=\lim_{h\to 0} \frac{(1-\cos h)(1+\cos h)}{h(1+\cos h)}=\lim_{h\to 0} \frac{1-\cos^2 h}{h(1+\cos h)}=\lim_{h\to 0}\frac{\sin h}{h}\lim_{h\to 0}\frac{\sin h}{1+\cos h}=1\cdot 0=0$$
A: The first one doesn't exist, because $\lim_{h\to 0}e^{-h}=1$. The second one can be done either via Taylor series, l'Hopital's rule or the the following trick:
$$\frac{\cos h-1}{h}=\frac{\cos h - \cos 0}{h-0}\to (\cos h )'|_{h=0}=0.$$
A: Use the reverse Prosthaphaeretic identities. 
$$\cos(a)-\cos(b)=-2\sin((a+b)/2)\sin((a-b)/2)$$ 
Taking into account that $\cos(0)=1$
Then you can write
$$\frac{\cosh-1}{h}=\frac{-2\sin(h/2)\sin(h/2)}{h}.$$
Use then that 
A: 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?
