Problems with limits at infinity within improper integrals

ninfinity
Messages
9
Reaction score
0

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}## ?
 
Physics news on Phys.org
hi ninfinity! :smile:
ninfinity said:
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/eb

what does eb/b -> ? :wink:
 
Isn't that ∞/∞? That doesn't make sense.
 
what does b3/b2 –> as b -> ∞ ? :wink:
 
ninfinity said:
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.
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top