limit and integration

**Sambit** · October 19th 2010, 11:42 PM

Evaluate ${\lim }\limits_{x \to 0} \frac{\int_0^x\,e^{t^2}(e^t-1)^2\, dt}{x(1-cos x)}$

**mr fantastic** · October 19th 2010, 11:43 PM

Originally Posted by Sambit

Evaluate ${\lim }\limits_{x \to 0} \frac{\int_0^x\,e^{t^2}(e^t-1)^2\, dt}{x(1-cos x)}$

Apply l'Hospital's Rule (and recall the Fundamental Theorem of Calculus when doing so).

**Also sprach Zarathustra** · October 20th 2010, 01:25 AM

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 $x\to 0$ :

$\frac{\int_0^0\,e^{t^2}(e^t-1)^2\, dt}{0(1-cos 0)}=\frac{0}{0}$

Next, ${x (1-cos(x))}' = {x sin(x)-cos(x)+1}\neq 0$ for every $x\neq0$

Applying l'Hospital's rule:
Now by the Fundamental Theorem of Calculus:

${\int_0^x\,e^{t^2}(e^t-1)^2\ dt}}'=e^{x^2}(e^x-1)^2$

${\lim }\limits_{x \to 0} {\frac{(\int_0^x\,e^{t^2}(e^t-1)^2\, dt)'}{(x(1-cos x))'}={\lim }\limits_{x \to 0} \frac{e^{x^2}(e^x-1)^2}{x sin(x)-cos(x)+1}$

The limit is still from the form of $\frac{0}{0}$ , applying l'Hospital's rule again...

**Sambit** · October 20th 2010, 09:22 AM

...............in this manner, if L'Hospitals rule is applied, at the $3rd$ oreder differentiation, we get:-
$\frac{((8x^3+24x^2+36x+20)e^{2x}+(-16x^3-24x^2-36x-14)e^x+8x^3+12x)e^{x^2}}{3cos x-xsin x}$

putting x=0, we get:- $6/3 = 2$

am i right?

**mr fantastic** · October 21st 2010, 03:31 AM

Originally Posted by Sambit

...............in this manner, if L'Hospitals rule is applied, at the $3rd$ oreder differentiation, we get:-
$\frac{((8x^3+24x^2+36x+20)e^{2x}+(-16x^3-24x^2-36x-14)e^x+8x^3+12x)e^{x^2}}{3cos x-xsin x}$

putting x=0, we get:- $6/3 = 2$

am i right?

No.

Here is the rest of post #3:

Originally Posted by Also sprach Zarathustra

${\lim }\limits_{x \to 0} \frac{e^{x^2}(e^x-1)^2)'}{(x sin(x)-cos(x)+1)'}={\lim }\limits_{x \to 0} \frac{2xe^{x^2}(e^x-1)^2+2e^{x^2}e^x(e^x-1)}{2sin(x)+xcos(x)}$

... l'Hospital's rule again...

${\lim }\limits_{x \to 0} \frac{(2xe^{x^2}(e^x-1)^2+2e^{x^2}e^x(e^x-1))'}{(2sin(x)+xcos(x))'}=2{\lim }\limits_{x \to 0} \frac{-(e^{x^2} (2 x^2+e^{2 x} (2 x^2+4 x+3)-e^x (4 x^2+4 x+3)+1)}{xsin(x)-3cos(x)}...=\frac{2}{3}$

**Sambit** · October 21st 2010, 05:45 AM

oh yes... actually my denominator was right, but i differentiated the numerator 4 times instead of 3 times....now i have got it correct...

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