How can I prove that $\lim_{x \rightarrow 5^-} \frac{1}{x-5}= - \infty$ using the epsilon and delta method. We know from the definition that: $\lim_{x \rightarrow 5^-} \frac{1}{x-5}= - \infty \Leftrightarrow $ For all $M>0$ there exists $\alpha > 0$, such that $0<|x-5|<\alpha \Rightarrow \frac{1}{x-5}< -M$.
I couldn't find alpha.
Thank you.
Work your way backwards. You want: $$\frac{1}{x-5}<-M$$ which is equivalent to: $$x-5 > -\frac{1}{M} \iff \color{blue}{5-x < \frac{1}{M}}$$ For the left-handed limit you're not looking at $x$-values satisfying $0<|x-5|<\alpha$ but (note that $|x-5|=5-x$ since $x<5$) you have: $0<\color{blue}{5-x<\alpha}$. Now spot a good choice for $\alpha$...