Why is L'Hôpital's rule giving the wrong answer? Given the following limit:
$$
 \lim_{x \rightarrow -1}  \frac{x^5+1}{x+1} 
$$
The solution using L'Hôpital's rule:
$$
 \lim_{x \rightarrow -1}  \frac{x^5+1}{x+1} = \begin{pmatrix} \frac{0}{0} \end{pmatrix} \rightarrow \lim_{x \rightarrow -1}  \frac{5x^4}{1} = 5 \cdot (-1)^4 = -5  
$$
This is wrong. How come?
EDIT: Well, a perfect example of a huge blunder.

As $\lim_{x\to-1}\frac{x^5+1}{x+1}$ is of the form $\frac00$
It's legal to apply L'Hospital's rule
$$\implies \lim_{x\to-1}\frac{x^5+1}{x+1}=\lim_{x\to-1}\frac{5x^4}{1}=5(-1)^4=5$$

Alternatively, 
as $x^5+1=(x+1)(x^4-x^3+x^2-x+1)$
If $x+1\ne0,$ $\displaystyle\frac{x^5+1}{x+1}=x^4-x^3+x^2-x+1$
If $x\to-1,x\ne-1\iff x+1\ne0$

Alternatively, 
we  can put $x+1=h\implies h\to0$ as $x\to-1$
