# Thread: Evaluate limit (with integral)?

1. ## Evaluate limit (with integral)?

Evaluate:

$\displaystyle \lim_{x\rightarrow 0} \frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}} dt$

2. $\displaystyle e^{\sqrt{t^2}}=e^{|t|}$

Since $\displaystyle 2-\sqrt{x}\leq t \leq 2+\sqrt{x}$ and x is close to 0, $\displaystyle t>0$ for the interval we're integrating in.

$\displaystyle \lim_{x\to 0}\frac{1}{\sqrt{x}}\int_{2-\sqrt{x}}^{2+\sqrt{x}}e^{|t|} dt=\lim_{x\to 0}\frac{1}{\sqrt{x}}\int_{2-\sqrt{x}}^{2+\sqrt{x}}e^{t} dt=$ $\displaystyle \lim_{x\to 0}\frac{1}{\sqrt{x}}(e^{2+\sqrt{x}}-e^{2-\sqrt{x}})=\{y=\sqrt{x},\ y\to 0, x\to 0 \}=\lim_{y\to 0}e^2\left(\frac{e^y-e^{-y}}{y}\right)=$ $\displaystyle \lim_{y\to 0}2e^2\left(\frac{\sinh y}{y} \right)$

At this point we can use the Maclaurin series for $\displaystyle \sinh x$:

$\displaystyle \sinh x=x+\frac{x^3}{3!}+O(x^5)$

I'll leave the rest to you.

3. The answer given in my text is simply $\displaystyle 2e^2$. Does that mean $\displaystyle \lim_{y\rightarrow 0} \frac{\sinh y}{y} = 1$ ?

EDIT: Yes! I forgot the Maclaurin Series.

4. Yes, you can see that when using the Maclaurin series since $\displaystyle \frac{1}{x}\left(x+\frac{x^3}{3!}+O(x^5) \right)=1+\frac{x^2}{3!}+O(x^4) \to 1, x\to 0$

5. We can also do it like this :
$\displaystyle \lim_{y\rightarrow 0} e^2\left(\frac{e^y - e^{-y}}{y}\right)\ \ \left(\frac{0}{0}\ \mbox{form}\right)$

$\displaystyle \mbox{Thus, by l'Hopital's rule,}$

Differentiating numerator and denominator,

$\displaystyle e^2\lim_{y\rightarrow 0} \left(\frac{e^y + e^{-y}}{1}\right)$
$\displaystyle = 2e^2$

6. Hi,
Originally Posted by fardeen_gen
Evaluate:

$\displaystyle \lim_{x\rightarrow 0} \frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}} dt$
A slightly different solution :

Let $\displaystyle F$ be an antiderivative of $\displaystyle t\mapsto \mathrm{e}^{\sqrt{t^2}}$. We have

\displaystyle \begin{aligned} \frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}}\ \mathrm{d}t &=\frac{F(2+\sqrt{x})-F(2-\sqrt{x})}{\sqrt{x}}\\ &=\frac{F(2+\sqrt{x})-F(2)}{\sqrt{x}}+\frac{F(2)-F(2-\sqrt{x})}{\sqrt{x}}\\ &=\frac{F(2+\sqrt{x})-F(2)}{\sqrt{x}}+\frac{F(2-\sqrt{x})-F(2)}{-\sqrt{x}} \end{aligned}
hence
\displaystyle \begin{aligned}\lim_{x\to 0}\frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}}\ \mathrm{d}t&= \lim_{h\to 0}\frac{F(2+h)-F(2)}{h}+\frac{F(2-h)-F(2)}{-h}\quad (\text{with}\ h=\sqrt{x})\\ &=2F'(2)\\ &=2\mathrm{e}^2 \end{aligned}

7. Leibnitz Rule:

If $\displaystyle F(x)=\int_{g(x)}^{h(x)}f(t)dt$
then
$\displaystyle F'(x)=f\left(h(x)\right)h'(x)-f\left(g(x)\right)g'(x)$

=$\displaystyle \lim_{x\to 0}\frac{\int_{2-\sqrt x}^{2+\sqrt x}e^{t^2}dt}{\sqrt x}$

$\displaystyle =\lim_{x\to 0}\frac{e^{\sqrt{(2+\sqrt x})^2}\frac{1}{2\sqrt x}-e^{\sqrt{(2-\sqrt x})^2}\frac{1}{-2\sqrt x}}{\frac{1}{2\sqrt x}}$

=$\displaystyle e^2+e^2$

=$\displaystyle 2e^2$