Solving Limits with L'Hopital's Rule

[Jeff12341234]

1. The limit as b approaches infinity always shows up as undefined on my calc so I don't know what to put for that section of the work.

2. What pat of the work is supposed to need L'hopital's rule? The integration?

 

[Dick]

1. The limit as b approaches infinity always shows up as undefined on my calc so I don't know what to put for that section of the work.

2. What pat of the work is supposed to need L'hopital's rule? The integration?

Same type of answer as with the other one. In e^(b(9-s)), 9-s is negative. That will go to zero as b->infinity. So will b*e^(b(9-s)) and b^2*e^(b(9-s)).

 

[Jeff12341234]

What about the 2nd question?

 

[Dick]

What about the 2nd question?

About l'Hopitals's? You can use it to prove something like b*exp(-kb) must go to zero as b->infinity if k>0. Write it as b/e^(kb) and use l'Hopital.

 

[Jeff12341234]

How does it specifically apply in this case? What am I supposed to need it for in this problem?

 

[Dick]

How does it specifically apply in this case? What am I supposed to need it for in this problem?

You want to show things like -b^2*e^(b(9-s))/(9-s) actually do go to zero when s>9 and b goes to infinity.

 

[Jeff12341234]

Ok, so it's supposed to be needed when evaluating the limits of each term correct?

 

[Dick]

Ok, so it's supposed to be needed when evaluating the limits of each term correct?

Yes, the parts you wrote '?' over. You'll want to show they actually do go to zero if s>9.

 

[Jeff12341234]

Did I do the work right? I get 0 in the denominator.. what's wrong here?

 

[Dick]

Did I do the work right? I get 0 in the denominator.. what's wrong here?

You can't use l'Hopital's rule on an expression unless it's indeterminate. Write it so it has the form 0/0 or infinity/infinity. Try b^2/e^b. That's infinity/infinity.

 

[Jeff12341234]

ok. So for that specific term that I tried to do, the limit looks like it would instead be infinity right? Am I allowed to do the limit of just -b^2 over s-9 by itself, along with each other term by itself, and the e^b(9-s) by itself, or do I have to expand it out and then do each term like I started to do and do the limit that way?

 

[Dick]

ok. So for that specific term that I tried to do, the limit looks like it would instead be infinity right? Am I allowed to do the limit of just -b^2 over s-9 by itself, along with each other term by itself, and the e^b(9-s) by itself, or do I have to expand it out and then do each term like I started to do and do the limit that way?

No, it's not infinity. It's zero. Limit b^2/e^b is infinity/infinity. Do l'Hopital. 2b/e^b. Still infinity/infinity. Do l'Hopital once more. Now it's not infinity/infinity. What is it?