Proving $f(x) = \frac{\sin x}{x}$ converges by improper integral test? So my professor talked about one example of improper integrals and I'm having difficulty understanding the general proof outline for proving convergence. 
I was given the problem to prove that $$f(x) = \begin{cases} \frac{\sin x}{x}, & x>0 \\ 1, & x=0 \end{cases}$$ converges
.
The proof then goes like (proof is in blockquotes and my question/comments follow underneath it):

$f$ is continuous at $x \ \forall \ x >0$ and $f$ is continuous at $x=0$.

My question: I know how to show continuity with the epsilon-delta argument, but other than that how can I tell if something is continuous? Specifically for this case, why do we have continuity for all $x>0$ and $x=0$? 

Therefore $f$ is RI on $[a,b] \ \forall \ b>0$. In particular, $f$ is RI on $[0,1]$. So $\displaystyle \int_0^{\infty} f(x) dx$ converges iff $\displaystyle \int_1^{\infty} f(x)dx$ converges.

I'm guessing that this is from the comparison test for improper integrals where if $|f(x)| \leq g(x)$ then $\displaystyle \int_a^{\infty} g(x) dx$ converges. Clearly the integral from $1$ to infinity is smaller than the integral from $0$ to infinity.

And so we look at
  \begin{align*}
\lim_{b \rightarrow \infty} \int_1^{b} \frac{\sin x}{x} dx &= \lim_{b \rightarrow \infty} \left[-\frac{1}{x} \cos x\right]_1^{b} - \int_1^{b} \frac{\cos x}{x^2} dx \\
&= \lim_{b \rightarrow \infty} \left(\frac{-\cos b}{b} +\cos(1)\right) - \int_1^{\infty} \frac{\cos x}{x^2} dx.
\end{align*}

I understand this is done by integration by parts. 

Finally, we have that $\displaystyle \int_1^{\infty} \frac{\cos x}{x^2} dx$ converges because $\displaystyle \left|\frac{\cos x}{x^2}\right| < \frac{1}{x^2}$ and $\displaystyle \int_1^{\infty} \frac{1}{x^2}$ converges.

Does $\displaystyle \int_1^{\infty} \frac{1}{x^2}$ converge from the fact that $\displaystyle \sum \frac{1}{n^p}$ converges if $p>1$? 
 A: 
QUESTION $1$:  "Specifically for this case, why do we have continuity for all x>0 and x=0?" 

For the Question $1$, we can easily show that 
$$\lim_{x\to 0}\frac{\sin(x)}{x}=1$$
whereby we conclude that $f$ is continuous at the origin.
To do so, we can use the result of THIS ANSWER in which I showed that 

$$\bbox[5px,border:2px solid #C0A000]{x\cos(x)\le \sin(x)\le x }\tag 1$$

for $0\le x\le \pi/2$.   Alternatively, we can apply L'Hospital's Rule.

Of course, if we know that $\sin(x)$ and $\frac1x$ are continuous for $x\ne 0$, then the quotient $\frac{\sin(x)}{x}$ is also continuous for $x\ne 0$.  And we are done!
To show that $\sin(x)$ is continuous, we can write for any $\epsilon>0$
$$\begin{align}
|\sin(x)-\sin(x_0)|&= 2\left|\cos\left(\frac{x+x_0}{2}\right)\sin\left(\frac{x-x_0}{2}\right)\right|\\\\
&\le 2\left|\sin\left(\frac{x-x_0}{2}\right)\right| \tag 2\\\\
&\le |x-x_0| \tag 3\\\\
&<\epsilon
\end{align}$$
whenever $|x-x_0|<\delta=\epsilon$.  Note that we used $(1)$ to go from $(2)$ to $(3)$.


QUESTION $2$:  "Does $\int_1^\infty \frac{1}{x^2}\,dx$ converge from the fact that $\sum_{n=1}^\infty \frac1{n^p}$ converges if $p>1$?" 

We can easily show that $\int_1^\infty \frac{1}{x^2}\,dx$ converges since
$$\lim_{L\to \infty}\int_1^L \frac{1}{x^2}\,dx=1-\frac{1}{L}\to 1$$
