Prove using the $\epsilon - \delta$ definition of limits that $\lim_{x\to3} \frac{5}{4x-11} = 5$.
I know how the setup should be given $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $|x-3| \lt \delta$ and $|\frac{5}{4x-11} - 5| \lt \epsilon$ but I can't do the computation to help me find $\delta$ can someone guide me in the right direction?
No comments:
Post a Comment