# If $\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$, compute: $\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$

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?

• It seems that the assumption that $f(x)$ is a polynomial is not required. Aug 4, 2019 at 6:05

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:)

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

• Why can we take $f(x)=3x-5$? Aug 4, 2019 at 7:36
• answer is in the OP's consideration. Aug 4, 2019 at 7:39
• My point is that we can't do it every time, because the limit might be dependent on the higher oder terms as well. Aug 4, 2019 at 7:52
• If so, then we have to follow the process described by Martund, otherwise it is a good way to solve this kind of problem. Aug 4, 2019 at 7:56
• 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. Aug 4, 2019 at 8:15

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 $$(\lim \frac {f(-x)+2} {(x+1)}) (\lim (f(-x)-2)) (\lim \frac 1 {x-1})=-6$$