Evaluating $\lim_{x~\to~ -1} \frac{x^{2n+1}+1}{x+1}$. What is the limit of: $$\lim_{x~\to~ -1} \frac{x^{2n+1}+1}{x+1}$$ 
Thanks
 A: Note: This solution is based on the original version of the problem, which had $x^{2n+1}$ not $x^{2n+1}+1$ in the numerator.
L'Hopital's rule doesn't apply here, as the numerator goes to $-1$ while the denominator goes to $0$.  The fraction blows up.  It is negative as $x\to -1+$ (from the right), and positive as $x\to -1-$ (from the left).  Hence the one-sided limits are $-\infty$ and $+\infty$, respectively.  The overall limit therefore does not exist.
A: On way to evaluate the limit is to write
$$\frac{x^{2n+1}+1}{x+1}=\sum_{k=0}^{2n}(-1)^{k}x^k\to 2n+1\,\,\text{as}\,\,x\to -1$$
A second way to evaluate the limit is to use L'Hospital's Rule.  We have 
$$\lim_{x\to -1}\frac{x^{2n+1}+1}{x+1}=\lim_{x\to -1}(2n+1)x^{2n}=2n+1$$
as expected!
A: Derivative of the function $x^{2n+1}$ at $x=-1$, which is $2n+1$.
A: Two ways to do this question:


*

*$$\lim\limits_{x \to -1}\dfrac{x^{2n+1}+1}{x+1} \overset{H}{=} \lim\limits_{x \to -1}\dfrac{(2n+1)x^{2n}}{1} = \left[2n+1\right](1)$$
because $(-1)^{\text{even}} = 1$. Hence we get $2n+1$.

*$$\lim\limits_{x \to -1}\dfrac{x^{2n+1}+1}{x+1} = \lim\limits_{x \to -1}\dfrac{x^{2n+1}-(-1)}{x-(-1)}\text{.}$$
Notice how similar this is to the definition of the derivative of a function $f$ evaluated at $x = -1$: $$f^{\prime}(-1) = \lim\limits_{x \to -1}\dfrac{x^{2n+1}-(-1)}{x-(-1)} = \lim\limits_{x \to -1}\dfrac{f(x)-f(-1)}{x-(-1)}$$
This implies that $f(x)= x^{2n+1}$. The derivative is given by $f^{\prime}(x) = (2n+1)x^{2n}$. Thus, $f^{\prime}(-1) = (2n+1)(1)=2n+1$.

