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

Question

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?

Answer 1

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:)

Answer 2

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

Answer 3

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