# Problems with limits at infinity within improper integrals

1. Jun 23, 2013

### ninfinity

1. The problem statement, all variables and given/known data

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

3. 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 $3e^{-2}$ ?

2. Jun 23, 2013

### tiny-tim

hi ninfinity!
you're asking whether be-b -> 0

that's b/eb

what does eb/b -> ?

3. Jun 23, 2013

### ninfinity

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

4. Jun 23, 2013

### tiny-tim

what does b3/b2 –> as b -> ∞ ?

5. Jun 23, 2013

### Staff: Mentor

∞/∞ 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.