I'm stuck with this problem:
Given $\lim \limits_{x \to 2} \frac{f(x+2)+3f(x^2)}{x^2+1}=3$ find $\lim \limits_{x \to 4} f(x)$
I've alreadly asked this question (with a different example) (Finding a limit inside a limit) and I understood it perfectly, but I'm stuck in the last step:
$g(x)=\frac{f(x+2)+3f(x^2)}{x^2+1}$ with $\lim \limits_{x \to 2} g(x)=3$
Solving for $f$:
$f(x+2)+f(x^2)=\frac{g(x)(x^2+1)}{5}$
Now finding the limit as ${x \to 2}$
$\lim \limits_{x\to 2} \frac{g(x)(x^2+1)}{5}=\lim \limits_{x\to 2}f(x+2)+f(x^2)=3$
And now I have no idea what to do…
$\lim \limits_{x\to 2} f(x+2) + \lim \limits_{x\to 2} f(x^2) = 3$
Best Answer
We do not reduce generality by assuming $f$ continuous at $x=2$ (we know that the limit exists), and the question can be simply solved as
$$\frac{f(4)+3f(4)}{2^2+1}=3\implies f(4)=\frac{15}4.$$