$\lim\limits_{x\rightarrow\infty} f(x)\ne L$ would mean that there is an $\epsilon>0$ such that for any $M>0$, there is an $x>M$ so that $|f(x)-L|\ge \epsilon$.
To use the above to show that $\lim\limits_{x\rightarrow\infty} f(x)$ does not exist, you would have to show that $\lim\limits_{x\rightarrow\infty} f(x)\ne L$ for any number $L$.
For your purposes, with $f(x)=x\sin x$, let $L$ be any number. We will show that $\lim\limits_{x\rightarrow\infty} f(x)\ne L$. Towards this end, take $\epsilon=1$. Now fix a value of $M$. Using Alex's answer, you can find an $x>M$ so that $|f(x)-L|\ge1$.
Thus $\lim\limits_{x\rightarrow\infty} f(x)$ does not exist.
(The limit might be infinite (it isn't, see Alex's answer again); but this is another matter...)
$$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$
For every $\epsilon > 0$, there exists a $\delta >0$ such that
$|x-2| < \delta \implies |(x^2+2x-7) - 1| < \epsilon$.
Very often, you solve these problems by looking at what $\epsilon$ needs to do and then working backwards to what $\delta$ needs to do. In this case
$$|(x^2+2x-7) - 1| = |x^2+2x-8| = |(x+4)(x-2)| = |x+4|\,|x-2|$$
So, we need to make $|x+4|\,|x-2| < \epsilon$. We know we are going to make $|x-2| < \delta$, but what do we do with $|x+4|$?
\begin{align}
|x-2| < \delta
&\implies 2-\delta < x < 2 + \delta \\
&\implies 4-\delta < x + 4 < 4 + \delta
\end{align}
The trick is to limit the size of $\delta$. There is no fixed limit that you need to use. Just pick one. I think $10$ is a nice round number so I am going to say, suppose $0 < \delta < 2$. Then
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0<\delta<2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (-2<-\delta<0)\\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
\end{align}
You should see that we now solve $10\delta < \epsilon$ for $\delta$. We get $\delta < \dfrac{\epsilon}{10}$. But wait! We made an assumption that $\delta < 2$. That's very easy to fix. Our final formula is
$\delta = \min\left\{2, \dfrac{\epsilon}{10} \right\}$.
Then we get our proof by adding one more line to the previous argument.
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0 < \delta < 2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (0<\delta<2) \\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
&\implies |(x^2+2x-7) - 1| < \epsilon
\end{align}
Best Answer
Let $f(x)=x^3+2x^2-x+1. $ Note $|f(x)-13|=|x-2||x^2+4x+7|$
For $0<|x-2|<1$, we have $1<x<3\Rightarrow12<x^2+4x+7<28$,
Let $\epsilon>0$, then take $\delta(\epsilon):=\min\{\epsilon/28,1\}$. For $0<|x-2|<\delta(\epsilon), $ we have
$$|f(x)-13|=|x-2||x^2+4x+7|<\frac{\epsilon}{28}28=\epsilon$$