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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?

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    $\begingroup$ 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? $\endgroup$ – John Omielan Jul 12 at 2:12
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    $\begingroup$ If you sub in $x = -1$, you get zero.The answer is... zero. So in this case the limit exists and is zero. $\endgroup$ – астон вілла олоф мэллбэрг Jul 12 at 2:13
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    $\begingroup$ But, shouldn't it be $f(-1)=-1$? $\endgroup$ – Sudix Jul 12 at 3:01
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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.

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