# Math Help - L'hospital rule

1. ## L'hospital rule

Problem: find values of a and b that satisfy the following

lim (sin x/x^3 + a + b/x^2) = 0
x->0

I did combine the whole function using common denominators

some one post this
Originally Posted by MarkFL2

$\lim_{x\to0^{+}}\frac{\sin(x)+ax^3+bx}{x^3}=0$

We have the indeterminate form 0/0, so application of L'Hôpital's rule gives:

$\lim_{x\to0^{+}}\frac{\cos(x)+3ax^2+b}{3x^2}=0$

In order for this limit to be bounded, we require the numerator to tend to 0, so we need $b=-1$, and we may write:

$\lim_{x\to0^{+}}\frac{\cos(x)+3ax^2-1}{3x^2}=0$

We have again the indeterminate form 0/0, so application of L'Hôpital's rule gives:

$\lim_{x\to0^{+}}\frac{-\sin(x)+6ax}{6x}=0$

We have again the indeterminate form 0/0, so application of L'Hôpital's rule gives:

$\lim_{x\to0^{+}}\frac{-\cos(x)+6a}{6}=0$

Can you now finish?
But why b= -1? does it has to be that exact number?

2. ## Re: L'hospital rule

If b is any other value, then the limit is no longer bounded. Do you see why?

3. ## Re: L'hospital rule

ok, i see it now

4. ## Re: L'hospital rule

another problem, probably still relate to L'hospital rule: lim x->0+ x^(x^2)

5. ## Re: L'hospital rule

I have another recommendation for getting the best help here, but only for your next question, no need to do this for this question:

Create a new topic for a new question. It is best to only post follow-up questions to the original problem in the same topic. I only mention these recommendations so that you will get the most prompt help.

We are given (assuming the limit exists):

$\lim_{x\to0^{+}}x^{x^2}=L$

Now, take the natural log of both sides, and use $\ln\left(\lim_{x\to c}f(x) \right)=\lim_{x\to c}\ln\left(f(x) \right)$, and using the properties of logs, see if you can get one of the forms for which L'Hôpital's rule may be applied.