# Limits Question

1. Sep 20, 2009

### JBD2

1. The problem statement, all variables and given/known data
$$Let $f(x)=\begin{cases}{} \frac{6a(x^2+1)}{2x^2+1}+\frac{b\log \big((x+1)^4\big)}{x} &\mbox{if } x<0,\\ -2a-4 &\mbox{if } x=0, \\ \frac{a\sin x}{x}+b &\mbox{if } x>0. \end{cases}$ Find $$a$$ and $$b$$ such that $$f$$ is continuous at $$x=0$$.$$

2. Relevant equations
Left and Right Hand limits are equal, as well as limit laws (If you're helping with the question I assume you know what they are)

3. The attempt at a solution
If f is continuous at x=0, the limit as x approaches 0 from the left should be equal to the limit as x approaches 0 from the right. With that, I took the limit of the part of f where x<0 and set it equal to the limit of the part of f where x>0.

Using L'hopital's rule I computed: $$\lim_{x \to 0-}\frac{4b\log{x+1}}{x}$$ and the result is 4b*log(e). If I set this equal to the right hand limit, I can calculate that $$b=\frac{-5a}{4\log{e}-1}$$. Now I am stuck and am not sure what to do. Thanks for the help.

2. Sep 20, 2009

### tiny-tim

Hi JBD2!

(I expect log here means log base e )

Now solve for b/a, use that to get rid of b, and make one of the limits equal to -2a - 4.