Proof using formal definition of limit that $ \lim_{x\to0} (x^2\sin(x^2+2x)+2)=2 $

Question

I'm having a hard time trying to understand how to prove the following, using the formal definition of limits:

$$ \lim_{x\to0} (x^2\sin(x^2+2x)+2)=2 $$

I already did the following the steps but I'm not sure if my thinking is correct:

$$ |x^2\sin(x^2+2x)+2-2|<\epsilon \implies |x^2\sin(x^2+2x)|<\epsilon \implies |x^2||\sin(x^2+2x)|<\epsilon $$

and assuming that $x\in{R}$ then $x^2|\sin(x^2+2x)|<\epsilon$

Now I also know that $-1<\sin(x)<1$ however I'm struggling to understand how to best apply this to the proof.

I'm not looking for a full solution to the problem, rather I'm interested in confirming if my thinking up until now is correct and maybe some hints to unblock the next steps.

Answer 1

You are almost there! Since the sine function is bounded by one and minus one, we have that \begin{align*} |x^{2}\sin(x^{2}+2x) + 2 - 2| = |x^{2}\sin(x^{2}+2)| \leq |x^{2}|\leq \varepsilon \Rightarrow |x|\leq\sqrt{\varepsilon} \end{align*}

Thus we conclude that for every $\varepsilon > 0$, there corresponds a $\delta = \sqrt{\varepsilon} > 0$ such that \begin{align*} |x - 0| \leq \delta = \sqrt{\varepsilon} \Rightarrow |x^{2}| \leq \varepsilon & \Rightarrow |x^{2}\sin(x^{2} + 2x)| \leq |\sin(x^{2}+2x)|\varepsilon \leq \varepsilon \end{align*} and we are done.

Hopefully this helps!

Answer 2

Assuming you are trying to find a $\delta > 0$ such that $|x - 0| < \delta$ implies the inequality you are working on. So you are trying to find an upper bound on $x$ in terms of $\varepsilon$.

You have more than $-1 \leq \sin x \leq 1$. You have $-1 \leq \sin(x^2 + 2x) \leq 1$. So $0 \leq |\sin(x^2 +2x)| < 1$. (The left-hand side of this is true for all moduli. The right-hand side comes from the previous inequality.) So you have $$ x^2 [\text{some number in $[0,1]$}] < \varepsilon \text{.} $$ The biggest the left-hand side can be is $x^2$ and this gives a bound on $|x|$ in terms of $\varepsilon$...