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Be $f(x)$ a polynomial in $\Bbb R$ such that: If$$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$$ Compute: $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$$

I noticed that, if the first limit exists, then $f(x)+2 = (x-1)P(x)$, where $P(x)$ is another polynomial.

Then $$\lim \limits_{x \to 1} P(x) = 3$$

I tried to use that in the second limit, but i can't proceed further.

Any hints?

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    $\begingroup$ It seems that the assumption that $f(x)$ is a polynomial is not required. $\endgroup$ – Teddy Aug 4 '19 at 6:05
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Observe that the denominator is going to zero in the first limit. Hence, the numerator must be going to zero for the limit to exist. Therefore, $f(1)=-2$. Now $$\lim_{x\to{-1}}\frac{(f(-x))^2-4}{x^2-1}$$ $$=\lim_{x\to{1}}\frac{(f(x))^2-4}{x^2-1}$$ $$=\lim_{x\to{1}}\frac{(f(x))+2}{x-1}×\lim_{x\to 1}\frac{f(x)-2}{x+1}$$ $$=3×\frac{-4}{2}$$ $$=-6$$ Hope it helps:)

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Given that $$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3\tag1$$ Take $~f(x)=3x-5~$, then the above limit is satisfied.

Now $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}=\lim \limits_{x \to -1} \frac{(-3x-5)^2-4}{x^2-1}$$ $$=\lim \limits_{x \to -1} \frac{(3x+3)(3x+7)}{x^2-1}$$ $$=\lim \limits_{x \to -1} \frac{3(3x+7)}{x-1}$$ $$=\frac{3(-3 +7)}{-1-1}=-6$$

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  • $\begingroup$ Why can we take $f(x)=3x-5$? $\endgroup$ – Botond Aug 4 '19 at 7:36
  • $\begingroup$ answer is in the OP's consideration. $\endgroup$ – nmasanta Aug 4 '19 at 7:39
  • $\begingroup$ My point is that we can't do it every time, because the limit might be dependent on the higher oder terms as well. $\endgroup$ – Botond Aug 4 '19 at 7:52
  • $\begingroup$ If so, then we have to follow the process described by Martund, otherwise it is a good way to solve this kind of problem. $\endgroup$ – nmasanta Aug 4 '19 at 7:56
  • $\begingroup$ In that case, that method would also fail, because we don't have information about the higher-order terms. I just wanted to point out that you can't do it every time, and you need to be careful. $\endgroup$ – Botond Aug 4 '19 at 8:15
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Hint: $\lim \frac {f(-x)^{2}-4} {x^{2}-1}= \lim \frac { {(f(-x)+2)} {(f(-x)-2)}} {(x-1) (x+1)}$ This is equal to $(\lim \frac {f(-x)+2} {(x+1)}) (\lim (f(-x)-2)) (\lim \frac 1 {x-1})=-6$

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