Proving that $\lim\limits_{x \to -1}(3x^2-3)\sin(x) = 0$.

Question

Prove that $\lim\limits_{x \to -1}(3x^2-3)\sin(x) = 0$.

So, by the definition I have to prove that

$$ \exists\delta>0 \text{ such that} $$ $$ 0<|x+1|<\delta \longrightarrow |(3x^2-3)\sin(x)|<\epsilon $$

What I did: $$ |(3x^2-3)\sin(x)|=3|x+1||x-1||\sin(x)|\leq3|x+1||x-1| $$

Let $\delta_1=1$ then:

$$ -1<x+1<1 \longrightarrow -3<x-1<-1<3 \longrightarrow |x-1|<3 $$

So: $$ 3|x+1||x-1||\sin(x)|\leq3|x+1||x-1|<3|x+1|\cdot3<\epsilon $$

$$ |x+1|<\frac\epsilon9 $$

So I let $\delta=\min(1,\frac\epsilon9)$ and I'm done?

Did I screw up anywhere?

What are other ways to prove this?

Answer 1

As it is, your proof is not complete. You've made it quite likely that $\delta=\min\left\{1,\dfrac\epsilon 9\right\}$ will suffice, but it's not a formal proof yet. To complete the formal proof, you have to show that this $\delta$ works for every $\epsilon>0$. This is quite tedious and routine, and it will be extremely similar to what you already did, however it must be done. Other than that the proof looks fine.

N.B. I would just like to note that the definition you wrote down, is missing something quite important. It should inculde at some point the phrase for every $\epsilon>0$. It's a petty comment, I know, but it realy is quite vital to the definition.

Answer 2

The basic result I use here is that, if $\lim_{x \to a} f(x) \ne 0$, then $\lim_{x \to a} f(x)g(x) = 0$ if and only if $\lim_{x \to a} g(x) = 0$.

First of all, $\sin(-1) \ne 0$, so $\lim\limits_{x \to -1}(3x^2-3)\sin(x) = 0$ if and only if $\lim\limits_{x \to -1}(3x^2-3) = 0$.

Second, $3x^2-3 =3(x^2-1) =3(x+1)(x-1) $, and $3((-1)-1) =-6$, so $\lim\limits_{x \to -1}(3x^2-3) = 0$ if and only if $\lim\limits_{x \to -1}(x+1) = 0$.

Finally, you be able to prove that $\lim\limits_{x \to -1}(x+1) = 0$.