Calculating $\lim_{x\to0}\frac{\int_{-2x}^x \sin(e^t)dt}{x}$ Hello everyone how can I calculate the limit of:
$\lim_{x\to0}\frac{\int_{-2x}^x \sin(e^t)dt}{x}$?
I tried to use L'hopital's rule and I got $\lim_{x\to0}\sin(e^x)-sin(e^{-2x})=sin(e^0)-sin(e^0)=0$
does it true?
 A: If $F(x)$ is an antiderivative of $\sin(e^x)$, that is $F'(x)=\sin(e^x)$, then
$$I(x)=\int_{-2x}^x\sin(e^t)\,dt=F(x)-F(-2x).$$
As $F(x)-F(-2x)\to F(0)-F(0)=0$ as $x\to0$, the Hospital can be used, giving
$$\lim_{x\to0}\frac{I(x)}x=\lim_{x\to0}\frac{I'(x)}1=\lim_{x\to0}(F'(x)+2F'(-2x))
=3F'(0)=3\sin1.$$
A: Option:
MVT for integrals.
$I(x) =(1/x)\int_{-2x}^{x}\ sin (e^t) dt=$
$(1/x)\sin (e^s) \int_{-2x}^{x}dt=$
$(1/x)\sin (e^s) 3x=$
$3\sin (e^s),$ where
$s \in (-2x, x)$ for  $x>0$, or $s \in (x, -2x)$ for $x <0. $
$\lim_{x \rightarrow 0} s =0;$
$\lim_{x \rightarrow 0}\sin (e^s) =\sin (1)$(Why?).
$\lim_{x \rightarrow 0}I(x)=3\sin (1)$.
A: You didn't apply the Leibniz Integral formula correctly.
$$\lim_{x\to0}\sin(e^x)+2\sin(e^{-2x})=\sin(e^0)+2\sin(e^0)=3\sin1$$
A: We have $\lim_{x\to 0} e^x=1$
Let let $\epsilon\gt 0$ be such that $0\lt 1-\epsilon\lt 1+\epsilon  \lt 3.14\simeq \pi$
Notice that $\sin x$ is positive on $(1-\epsilon,1+\epsilon)$
Then for this $\epsilon ,\exists \delta \gt 0$ such that $\forall x \in (0, \delta/2), e^x \in (1-\epsilon, 1+\epsilon) $
Also $\sin x\le x, \forall x\ge 0$
So we are able to write the following inequality for $x$ positive and close enough to $0$
$0\le \int_{-2x}^{x} sin(e^t)dt \le \int_{-2x}^{x} e^t=e^x-e^{-2x} $
Thus the given integral approaches to zero as $x\to 0$ by Sandwich Theorem
So given limit can be evaluated using L Hospitals Rule.
Again By Leibnitz Rule
Now  $\frac {d}{dx} \int_{-2x}^{x} sin(e^t)dt=1sin(e^x)-(-2)sin(e^{-2x})$
Taking limit $x\to 0$ above , the required limit is $3sin(1)$
My intention here is to present the reason why L Hospitals rule can be applied rather than actual computation.
A: Rewrite the expression under limit as $$\frac{1}{x}\int_{0}^{x}f(t)\,dt+2\cdot \frac{1}{(-2x)}\int_{0}^{-2x}f(t)\,dt$$ where $f(t) =\sin(e^t) $. Using Fundamental Theorem of Calculus the above tends to $$f(0)+2f(0)=3\sin 1$$ as $x\to 0$.
When dealing with limit of integrals do think of Fundamental Theorem of Calculus. In many cases this will help to get the answer easily. Also notice that L'Hospital's Rule can be avoided here.
