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