Finding a limit inside a limit 2

Question

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...

$\lim \limits_{x\to 2} f(x+2) + \lim \limits_{x\to 2} f(x^2) = 3$

Answer 1

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

$$\frac{f(4)+3f(4)}{2^2+1}=3\implies f(4)=\frac{15}4.$$

Answer 2

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

Answer 3

You are almost there!

First, let's note that since $2^2+1=5$, so $\lim\limits_{x\to2}f(x+2)+3\lim_{x\to2}f(x^2)=15$, not $3$.

Next, note a very special property of $2$: $2^2=2+2$. This is important because we know then that $\lim\limits_{x\to2}f(x+2)=\lim\limits_{x\to2}f(x^2)=\lim\limits_{x\to4}f(x)$ as polynomials are continuous functions. Hence, $4\cdot\lim\limits_{x\to4}f(x)=15$ so $$\lim_{x\to4}f(x)=\color{red}{\frac{15}4}$$

Answer 4

$$\lim \limits_{x \to 2} \frac{f(x+2)+3f(x^2)}{x^2+1}=3$$

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 _{x\to 4} f(x))= 15 $ which results in $\lim _{x\to 4}=15/4$