Solving a Limit Problem if the Limit Exists

My question is, consider $$\lim [(x^{2} +2x +1) / (x^{4} -1)]$$ for $$x \to -1$$ (just arrow sign to the right)

What I have done so far is

$$(x+1)^{2}/[(x^{2} +1)(x+1)(x-1)]$$

$$(x+1)/[(x^{2}+1)(x-1)]$$

I'm not too sure what to do at this point. If I substitute in $$x=-1$$, I would get $$0$$.

So in this case would the limit not exist?

• Welcome to MSE. Since you simplified the expression and found it had a definite value of $0$ at $x = -1$, why do you think the limit doesn't exist in this case? – John Omielan Jul 12 at 2:12
• If you sub in $x = -1$, you get zero.The answer is... zero. So in this case the limit exists and is zero. – астон вілла олоф мэллбэрг Jul 12 at 2:13
• But, shouldn't it be $f(-1)=-1$? – Sudix Jul 12 at 3:01

Note that $$x^2+2x+1=(x+1)^2$$ and $$x^4-1=(x^2+1)(x+1)(x-1),$$ so that $$\frac{x^2+2x+1}{(x^2+1)(x+1)(x-1)} = \frac{x+1}{x^2+1)(x-1)},$$ so $$\lim_{x \to -1} \frac{x^2+2x+1}{(x^2+1)(x+1)(x-1)} = \frac{0}{2 \cdot -2} = 0.$$
$$0$$ is a perfectly fine answer. You may be confusing this with getting 0 in the denominator, which either means the limit doesn't exist or more work needs to be done.