Problems with limits at infinity within improper integrals

[ninfinity]

Homework Statement

$\int_{2}^{\infty} ue^{-u} du$

The Attempt at a Solution

What I did was find the family of functions described by the indefinite integral $\int ue^{-u} du$ then found the limit as b increases without bound. $$=\lim_{b\rightarrow \infty} -ue^{-u}-e^{-u}\mid_{2}^{b}$$ $$=\lim_{b\rightarrow \infty} \Big(-be^{-b}-e^{-b}\Big)~ -~ \Big(-2e^{-2}-e^{-2}\Big)$$ $$=\lim_{b\rightarrow \infty} e^{-b} \Big(-b-1\Big) + 3e^{-2}$$

My problem is trying to remember when $e^{-b}$ is approaching 0 and $\Big(-b-1\Big)$ is getting infinitely more negative...do they "cancel" each other out? Does that part of the limit approach 0 instead of ∞? And then does that mean that the definite integral converges at $3e^{-2}$ ?

 

[tiny-tim]

hi ninfinity!

My problem is trying to remember when $e^{-b}$ is approaching 0 and $\Big(-b-1\Big)$ is getting infinitely more negative...do they "cancel" each other out? Does that part of the limit approach 0 instead of ∞? And then does that mean that the definite integral converges at $3e^{-2}$ ?

you're asking whether be^-b -> 0

that's b/e^b …

what does e^b/b -> ? ​

 

[ninfinity]

Isn't that ∞/∞? That doesn't make sense.

 

[tiny-tim]

what does b³/b² –> as b -> ∞ ?

 

[Mark44]

Isn't that ∞/∞? That doesn't make sense.

∞/∞ isn't a number; it's an indeterminate form, as is 0/0 and several others. If you take a limit and get one of these indeterminate forms, it means you aren't done yet.