Limits as x appraoches infinity

[Idealistic]

Why is the limit of integral (from 0 to infinity) of x/(x^2 + 2)^2 = 1/4? I took the antiderivative and got:

-1/(2(x^2 + 2)) and as x approaches infinty I figured it would be negative one divided by 2(infinity + 2) which would approach zero?

 

[galactus]

Try it this way:

\(\displaystyle \int_{0}^{L}\frac{x}{(x^{2}+2)^{2}}=\frac{1}{4}-\frac{1}{2(L^{2}+2)}\)

Now, take the limit of that as x-->inf.

Now, you can see why it is 1/4.

 

[Idealistic]

Try it this way:

\(\displaystyle \int_{0}^{L}\frac{x}{(x^{2}+2)^{2}}=\frac{1}{4}-\frac{1}{2(L^{2}+2)}\)

Now, take the limit of that as x-->inf.

Now, you can see why it is 1/4.

I can see in that equation why the limit would be 1/4, but if the antiderivative of the original equation \(\displaystyle \frac{x}{(x^{2}+2)^{2}}}\), is \(\displaystyle \frac{1}{2(x^{2}+2)}\), how did you manage to separate the anti derivative and 1/4? I mean, subtract 1/4 from the antiderivative - which appears to approach zero.

 

[galactus]

The antiderivative is indeed \(\displaystyle \frac{-1}{2(x^{2}+2)}\)

Plug in 0 and we get -1/4.

Plug in L and we get \(\displaystyle \frac{-1}{2(L^{2}+2)}\)

By the first fundamental theorem of calcarooney, subtract them:

\(\displaystyle \frac{-1}{2(L^{2}+2)}-(\frac{-1}{4})=\frac{1}{4}-\frac{1}{2(L^{2}+2)}\)

You see now, don't you?. Now, take the limit as L heads toward infinity and you can see it's 1/4 because the one with the L term goes to 0.

 

[Idealistic]

Yes, I see now.

Thanks.