Evaluate $\displaystyle {\lim }\limits_{x \to 0} \frac{\int_0^x\,e^{t^2}(e^t-1)^2\, dt}{x(1-cos x)}$
Hi, I will continue mr fantastic idea...
Why you can apply l'Hospital's rule?
Let we see what happens to the whole expession when $\displaystyle x\to 0$:
$\displaystyle \frac{\int_0^0\,e^{t^2}(e^t-1)^2\, dt}{0(1-cos 0)}=\frac{0}{0}$
Next, $\displaystyle {x (1-cos(x))}' = {x sin(x)-cos(x)+1}\neq 0$ for every $\displaystyle x\neq0$
Applying l'Hospital's rule:
Now by the Fundamental Theorem of Calculus:
$\displaystyle {\int_0^x\,e^{t^2}(e^t-1)^2\ dt}}'=e^{x^2}(e^x-1)^2$
$\displaystyle {\lim }\limits_{x \to 0} {\frac{(\int_0^x\,e^{t^2}(e^t-1)^2\, dt)'}{(x(1-cos x))'}={\lim }\limits_{x \to 0} \frac{e^{x^2}(e^x-1)^2}{x sin(x)-cos(x)+1}$
The limit is still from the form of $\displaystyle \frac{0}{0}$, applying l'Hospital's rule again...
...............in this manner, if L'Hospitals rule is applied, at the $\displaystyle 3rd$ oreder differentiation, we get:-
$\displaystyle \frac{((8x^3+24x^2+36x+20)e^{2x}+(-16x^3-24x^2-36x-14)e^x+8x^3+12x)e^{x^2}}{3cos x-xsin x}$
putting x=0, we get:- $\displaystyle 6/3 = 2$
am i right?