How to solve a limit that can be factored but doesn't help?

Question

I saw examples that can be factored, eliminating the part that causes the indetermination, none of this type. The other option is by rationalize but dont know how to apply it here.

$$\lim_{x \to 4} \frac{2x^2+7x+5}{x^2-16}$$ I tried by factoring, doesn't help $$\lim_{x \rightarrow 4} \frac{(2x+5)(x+1)}{(x-4)(x+4)} \\$$

UPDATE: I make a mistake, the numerator is ${2x^2+7x+5}$ not ${x^2+7x+5}$, really sorry.

To make life easier, let $x=4+y$. It gives $\frac{y^2+15 y+49}{y^2+8 y}$ and now $y\to 0$ — Claude Leibovici, Sep 01 '21 at 15:30
Think about the values of the numerator and the denominator when $x$ is very close to 4, say $4.01$. — Andreas Blass, Sep 01 '21 at 15:31
Your factorization for the $x^2+7x+5$ term seems not correct, even if it is not the main issue with the resolution. — user, Sep 01 '21 at 15:37
@user You're right sorry, I updated, it should be $2x^2+7x+5$ — user2495207, Sep 01 '21 at 15:53
@user2495207 Thanks, in any case this is not the main issue for the resolution. — user, Sep 01 '21 at 15:59

user · Answer 1 · 2021-09-01T16:51:00.857

4

HINT

Your limit is in the form

$$\lim_{x \rightarrow 4} \frac{2x^2+7x+5}{(x-4)(x+4)} = \lim_{x \rightarrow 4} \frac{(2x+5)(x+1)}{(x+4)}\cdot \lim_{x \rightarrow 4} \frac{1}{x-4}$$

with

$$\lim_{x \rightarrow 4} \frac{(2x+5)(x+1)}{(x+4)} =\frac{65}{8}$$

then all boils down in that one

$$\lim_{x \rightarrow 4} \frac{1}{x-4} $$

edited Sep 01 '21 at 16:51

answered Sep 01 '21 at 15:32

user

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How to solve a limit that can be factored but doesn't help?

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