<s>[INST] Limits at negative infinity Prove that $\lim_{x\to -\infty}$ $\frac{x}{x+2}$ = 1.
Please help me check if this is correct:


Let $\varepsilon > 0$ be given. Choose $M=2-\frac{2}{\varepsilon}$
$$\begin{align}
  x<M \Rightarrow x-2 < M-2\\
  \frac{1}{x-2} > \frac{1}{M-2}\\
  |f(x)-1|=|\frac{x}{x-2}-1|=|\frac{2}{x-2}|\\
  <|\frac{2}{M-2}|=\frac{-2}{M-2}=\varepsilon
  \end{align}$$


 [/INST] Since $x \to -\infty$, we may assume $M < 0$.  Then your line $\frac{1}{x-2} > \frac{1}{M-2}$ is a relationship between two negative quantities.  Thus $\left|\frac{1}{x-2}\right| < \left|\frac{1}{M-2}\right|$.  Can you take it from there?
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