Exploring Limits: Refreshingly Simple!

[chwala]

Homework Statement

see attached.

Relevant Equations

understanding of the l'hopital rule

Refreshing..i will attempt part (a) first ...of course this is easy...
$$\displaystyle{\lim_{x \to \infty}}\frac{x^2}{e^x} $$
$$\displaystyle{\lim_{x \to \infty}}\frac{2x}{e^x} $$
$$\displaystyle{\lim_{x \to \infty}}\frac{2}{e^x} =0$$

 

[chwala]

Now to part (b);

$\int_0^∞ x^2e^{-x} dx$=${\lim_{t \to \infty}}$$\int_0^t x^2e^{-x} dx$
having indicated this then we shall have;

$\int x^2e^{-x} dx$$=\dfrac{-x^2}{e^x}$+$\int 2xe^{-x} dx$
=$\dfrac{-x^2}{e^x}-2xe^{-x}$+$\int e^{-x} dx$
=$\dfrac{-x^2}{e^x}-2xe^{-x}-e^{-x}$
=$\dfrac{-x^2}{e^x}-\dfrac{2x}{e^x}-\dfrac{1}{e^x}$
=${\lim_{t \to \infty}}$$\dfrac{-t^2}{e^t}-\dfrac{2t}{e^t}-\dfrac{1}{e^t}-[0-0-1]$ =${\lim_{t \to \infty}}$$\dfrac{-2t}{e^t}-\dfrac{2}{e^t}-\dfrac{1}{e^t}-[0-0-1]$
$={\lim_{t \to \infty}}$$\dfrac{-2}{e^t}-\dfrac{2}{e^t}-\dfrac{1}{e^t}-[0-0-1]=[0-0-1]-[0-0-1]=-1+1=0$
thus converges and its value is $0$.

your thoughts guys

 

[DrClaude]

Have you looked at the plot for $x^2 e^{-x}$. Would it make sense that the integral would be 0?

 

[chwala]

Have you looked at the plot for $x^2 e^{-x}$. Would it make sense that the integral would be 0?

Let me check...

 

[chwala]

Let me amend my post; it ought to be

Now to part (b);

$\int_0^∞ x^2e^{-x} dx$=${\lim_{t \to \infty}}$$\int_0^t x^2e^{-x} dx$
having indicated this then we shall have;

$\int x^2e^{-x} dx$$=\dfrac{-x^2}{e^x}$+$\int 2xe^{-x} dx$
=$\dfrac{-x^2}{e^x}-2xe^{-x}$+$2\int e^{-x} dx$
=$\dfrac{-x^2}{e^x}-2xe^{-x}-2e^{-x}$
=$\dfrac{-x^2}{e^x}-\dfrac{2x}{e^x}-\dfrac{2}{e^x}$...therefore on taking limits as required we shall have;

=${\lim_{t \to \infty}}$$\left[\dfrac{-t^2}{e^t}-\dfrac{2t}{e^t}-\dfrac{2}{e^t}\right]$$-[0-0-2]$
${\lim_{t \to \infty}}$$\left[\dfrac{-2t}{e^t}-\dfrac{2}{e^t}-\dfrac{2}{e^t}\right]$$-[-0-0-2]$

$={\lim_{t \to \infty}}$$\left[\dfrac{-2}{e^t}-\dfrac{2}{e^t}-\dfrac{2}{e^t}\right]$$-[0-0-2]=[-0-0-0]-[-0-0-2]=0+2=2$

thus converges and its value is $2$.

your thoughts guys

 

[fresh_42]

Let me amend my post; it ought to be

Now to part (b);

$\int_0^∞ x^2e^{-x} dx$=${\lim_{t \to \infty}}$$\int_0^t x^2e^{-x} dx$
having indicated this then we shall have;

$\int x^2e^{-x} dx$$=\dfrac{-x^2}{e^x}$+$\int 2xe^{-x} dx$
=$\dfrac{-x^2}{e^x}-2xe^{-x}$+$2\int e^{-x} dx$
=$\dfrac{-x^2}{e^x}-2xe^{-x}-2e^{-x}$
=$\dfrac{-x^2}{e^x}-\dfrac{2x}{e^x}-\dfrac{2}{e^x}$...therefore on taking limits as required we shall have;

=${\lim_{t \to \infty}}$$\left[\dfrac{-t^2}{e^t}-\dfrac{2t}{e^t}-\dfrac{2}{e^t}\right]$$-[0-0-1]$
${\lim_{t \to \infty}}$$\left[\dfrac{-2t}{e^t}-\dfrac{2}{e^t}-\dfrac{2}{e^t}\right]$$-[-0-0-2]$

$={\lim_{t \to \infty}}$$\left[\dfrac{-2}{e^t}-\dfrac{2}{e^t}-\dfrac{2}{e^t}\right]$$-[0-0-2]=[-0-0-0]-[-0-0-2]=0+2=2$

thus converges and its value is $2$.

your thoughts guys

I checked the result, which is correct. So either you made a mistake twice in different directions, or it is ok.

 

[chwala]

I checked the result, which is correct. So either you made a mistake twice in different directions, or it is ok.

I had made a mistake earlier by missing the $'2'$ ...after integration by parts...i.e in my post $2$ ...I amended that in my post $5$...I didn't want to interfere with the post so as to make it easier for people to follow...
Just amended post $5$ ...latex typo...

 

[nuuskur]

The integrand is non-negative, so the integral can be zero only if the integrand is zero almost everywhere. Clearly, that is not the case. The result you arrived at later is correct.