Find $\lim_\limits{x\to 1}{\frac{f(x)-2\cdot x^2}{x-1}}$

Question

Let $f: \mathbb{R}\mapsto \mathbb{R}$ be a function such that: $$\lim_\limits{x\to 1}{\frac{f^2(x)+f(x)-6}{x-1}}=5$$ If we know that $$\lim_\limits{x\to 1}{\frac{f(x)-2\cdot x^2}{x-1}}=a\in \mathbb{R}$$ find the value of 'a' without using derivatives.

How can we solve this?

I see that we can easily factor the $f^2(x)+f(x)-6=(f(x)+3)\cdot (f(x)-2)$ but I don't understand how to move on...

Answer 1

Hint

$$a=\lim_\limits{x\to 1}{\frac{f(x)-2\cdot x^2}{x-1}}=\lim_\limits{x\to 1}{\frac{f(x)-2}{x-1}}-2\lim_\limits{x\to 1}{\frac{x^2-1}{x-1}}=\lim_\limits{x\to 1}{\frac{f(x)-2}{x-1}}-4$$

This tells you that $$a+4=\lim_\limits{x\to 1}{\frac{f(x)-2}{x-1}}$$

Now write $$f^2(x)+f(x)-6=(f(x)-2)(f(x)+3)$$ and use the above twice: once to get rid of the fraction and second time to deduce that $\lim_{x \to 1}f(x)=2$.

Answer 2

Maybe this is not the shortest path, but I find it less "dangerous":

First, we shall prove that $f$ has a limit when $x\to 1$.

From the equality $$\lim_{x\to1}\frac{f(x)-2x^2}{x-1}=a$$ we can say that for any $\epsilon>0$ there exists some $\delta\in(0,\frac\epsilon{|a|+\epsilon})$ such that $$0<|x-1|<\delta\implies\left|\frac{f(x)-2x^2}{x-1}-a\right|<\epsilon$$ or $$0<|x-1|<\delta\implies|f(x)-2x^2|<|x-1|(|a|+\epsilon)<\epsilon$$ Therefore $\lim_{x\to 1}(f(x)-2x^2)=0$. Hence, $\lim_{x\to 1}f(x)$ does exist and is $2$. To show this, just put to work the $\epsilon-\delta$ machine together with the inequality $|f(x)-2|\le|f(x)-2x^2|+|2x^2-2|$.

Now we use the first limit to calculate the second one.

The polynomial $X^2+X-6$ is reducible. In fact, it is $(X-2)(X+3)$. Then $$\frac{f(x)^2+f(x)-6}{x-1}=\frac{(f(x)-2)(f(x)+3)}{x-1}$$

Furthermore, $$\frac{f(x)-2x^2}{x-1}=\frac{f(x)-2+2-2x^2}{x-1}=\frac{f(x)-2}{x-1}-2(x+1)$$ hence $$\lim_{x\to 1}\frac{f(x)-2}{x-1}=\lim_{x\to 1}\left(\frac{f(x)-2x^2}{x-1}+2x+2\right)=a+4$$

Finally $$\lim_{x\to 1}\frac{f(x)^2+f(x)-6}{x-1}=5(a+4)$$ Thus, $a+4=1$ and $a=-3$.