A question from Introduction to Analysis by Arthur Mattuck:
Prove that $\Gamma (x)=\int_0^\infty t^{x-1}e^{-t}dt$ is continuous at $x=1^+$.
(Method: consider $|\Gamma(1+h)-\Gamma(1)|.$ To estimate it, break up the interval $[0,\infty)$ into two parts. Remember that you can estimate differences of the form $|f(a)-f(b)|$ by using the Mean-value Theorem, if $f(x)$ is differentiable on the relevant interval.
$|\Gamma(1+h)-\Gamma(1)|=\int_0^\infty (t^h-1)e^{-t}dt$. I don't know how to apply the Mean-value Theorem to it. I haven't learned differentiating under the integral sign.
For any $h>0$ $$\int_{0}^{+\infty}(t^h-1)e^{-t}\,dt\stackrel{t\mapsto-\log x}{=}-1+\int_{0}^{1}\left(-\log x\right)^h\,dx \tag{1}$$ but for any $x\in(0,1)$ $$0<-\log x=\int_{x}^{1}\frac{du}{u}\stackrel{\text{Cauchy-Schwarz}}{\leq}\sqrt{\int_{x}^{1}du\int_{x}^{1}\frac{du}{u^2}}=\frac{1}{\sqrt{x}}-\sqrt{x} \tag{2}$$ and by assuming $h\in(0,1)$ $$\begin{eqnarray*} \int_{0}^{+\infty}(t^h-1)e^{-t}\,dt &\leq& -1+\int_{0}^{1}\left(\frac{1}{\sqrt{x}}-\sqrt{x}\right)^h\,dx\\&=&-1+\int_{0}^{1}(1-x)^h x^{-h/2}\,dx\\&\leq&-1+\sqrt{\int_{0}^{1}(1-x)^{2h}\,dx\int_{0}^{1}x^{-h}\,dx} \\&=&-1+\sqrt{\frac{1}{(1+2h)(1-h)}}\tag{3}\end{eqnarray*}$$ On the other hand $$ \int_{0}^{+\infty}(t^h-1)e^{-t}\,dt \geq \int_{0}^{1}(t^h-1)e^{-t}\,dt $$ and $$0\leq \int_{0}^{1}(1-t^h)e^{-t}\,dt\leq \int_{0}^{1}(1-t^h)\,dt = \frac{h}{h+1} $$ imply $$ -\frac{h}{h+1}\leq\int_{0}^{+\infty}(t^h-1)e^{-t}\,dt \leq -1+\frac{1}{\sqrt{(1+2h)(1-h)}}\tag{4}$$ and the claim follows by squeezing.