How to show $\lim_{x \rightarrow 0} ~\frac{\int_{0}^{x^2} e^{\sqrt{1+t}} dt}{x^2}=e?$

Question

How to show $\lim_{x \rightarrow 0} ~\frac{\int_{0}^{x^2} e^{\sqrt{1+t}} dt}{x^2}=e?$

If i substitute $y=x ^2$, as $x \rightarrow 0 \implies y\rightarrow 0+$, so the limit is same as $\lim_{y \rightarrow 0+} ~\frac{\int_{0}^{y} e^{\sqrt{1+t}} dt}{y}$. Now i know i have to substitute $F(y)= \int_{0}^{y} e^{\sqrt{1+t}} dt$, but i cannot proceed further(i am noob). Can you please help me with this?

Answer 1

By the Fundamental Theorem of Calculus,\begin{align}\lim_{y\to0}\frac{\int_0^ye^{\sqrt{1+t}}}y\,\mathrm dt&=\left.\frac{\mathrm d}{\mathrm dy}\int_0^ye^{\sqrt{1+t}}\,\mathrm dt\right|_{y=0}\\&=e^{\sqrt{0+1}}\\&=e.\end{align}

Answer 2

It is 0/0 form so use l'hopital. Using Newton-Leibnitz rule, it becomes $\frac {e^{\sqrt{1+x^2}}\frac {d(x^2)}{dx}-{e^{\sqrt{1+0}}\frac {d(0)}{dx}} }{2x}=\frac {e^{\sqrt{1+x^2}}2x}{2x}$. Then substitute x=0

Answer 3

We can use L'Hospital rule here.

Derivative of numerator (Using Lebinitz rule):

$$e^{\sqrt{1+x^2}} 2x$$

Derivative of denominator:

$$2x$$

Therefore the limit comes down to:

$$\lim_{x\to0} \frac{e^{\sqrt{1+x^2}} 2x}{2x}$$

Answer 4

One way is to recognize that the expression is equal to the derivative (of the integral). Then apply the fundamental theorem of calculus.

Apparently, L'Hôpital's rule may also be used (in conjunction with FTC).

Answer 5

Elementary properties of the integral give you the following: whenever $f$ is a continuous function on an interval $(a,b)$ and $c \in (a,b)$, then $$\lim_{y \to 0} \frac 1y\int_c^{c+y} f(t) \, dt = f(c).$$

Take $(a,b) = (-1,\infty)$, $c = 0$, and $f(t) = e^{\sqrt{1+t}}$.