Question

Help Pleaseee

Answer 1

\[\int\limits_{1}^{\infty} \frac{ 2+e^{-x} }{ x } dx\]

Use the Comparison Theorem to Determine whether is is converging or diverging.

Answer 2

Isnt it diverging? Because \[e^{-\infty}\] approaches zero faster than x approaches infinity. @agent0smith

Answer 3

As x approaches infinity, e^-x approaches zero... which means the function behaves more like 2/x

You could compare to 1/x, since your function will always be larger than it (you have 2 on top, AND you're adding e^-x which is always a positive number)

Answer 4

Oops i meant to write converging. So it is converging right?

Answer 5

No... because integral 1/x diverges, and your function, as i described above, is even larger.

Answer 6

\[\Large \frac{ 1 }{ x } < \frac{ 2 }{ x } < \frac{ 2+e^{-x} }{ x }\](at least for the given limits of the integral)

Answer 7

And you know the integral of 1/x for the same limits diverges. Anything larger also diverges.

Answer 8

Yea but 1/x
is small over big,
isn't that approaching 0?

Answer 9

Yes, the function approaches zero; that doesn't mean the integral of the function does.\[\large \int\limits_{1}^{\infty} \frac{ 1 }{ x } dx\]find this and you'll see it does not converge.

Answer 10

Oh I see. Okay