Vtyjkyuk

I'm stuck with this problem:

Given $lim limits_x to 2 fracf(x+2)+3f(x^2)x^2+1=3$ 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:

$g(x)=fracf(x+2)+3f(x^2)x^2+1$ with $lim limits_x to 2 g(x)=3$

Solving for $f$:

$f(x+2)+f(x^2)=fracg(x)(x^2+1)5$

Now finding the limit as $x to 2$

$lim limits_xto 2 fracg(x)(x^2+1)5=lim limits_xto 2f(x+2)+f(x^2)=3$

And now I have no idea what to do...

$lim limits_xto 2 f(x+2) + lim limits_xto 2 f(x^2) = 3$

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

$$fracf(4)+3f(4)2^2+1=3implies f(4)=frac154.$$

$$lim limits_x to 2 fracf(x+2)+3f(x^2)x^2+1=3$$

Assuming the limits exist, we have $$lim limits_x to 2 f(x+2)+3f(x^2) = 3(2^2)+1=15$$

Thus $4 (lim _xto 4 f(x))= 15 $ which results in $lim _xto 4=15/4$

For the first limit, do the variable change $u = x+2$, so that $u to 4$ as $x to 2$, giving
$$ lim_xto 2 f(x+2) =lim_uto 2 f(u), $$
and for the second limit, do the variable change $v = x^2$, so that $v to 4$ as $x to 2$, giving
$$ lim_xto 2 f(x^2)=lim_vto 2 f(v), $$
so that the LHS of the final equality in your post is $$2lim_xto 4 f(x). $$ Also, the RHS in that same equality is wrong, as pointed out by the others.

You are almost there!

First, let's note that since $2^2+1=5$, so $limlimits_xto2f(x+2)+3lim_xto2f(x^2)=15$, not $3$.

Next, note a very special property of $2$: $2^2=2+2$. This is important because we know then that $limlimits_xto2f(x+2)=limlimits_xto2f(x^2)=limlimits_xto4f(x)$ as polynomials are continuous functions. Hence, $4cdotlimlimits_xto4f(x)=15$ so $$lim_xto4f(x)=colorredfrac154$$