Finding a and b for Continuous f(x) at x=0

[JBD2]

Homework Statement

$$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 \(x=0\).$$

Homework 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)

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.

 

[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.