Proving $\lim_{x \to a} \frac{x}{1+x} = \frac{a}{1+a}$

Question

I am trying to prove that, for all $a \in \mathbb{R}, a \neq -1$,

$$\lim_{x \to a} \frac{x}{1+x} = \frac{a}{1+a}$$

by using the $\epsilon - \delta$ formalism. I'm completely lost and unsure on how to start.

Answer 1

First we observe that, if $\lim_{x \to a}\frac{x}{1+x} = \frac{a}{1+a}$ or $\lim_{\xi \to 1+a}\frac{\xi - 1}{\xi} = \frac{a}{1+a}$, then $$ \lim_{x \to a}\frac{x}{1+x} = \lim_{\xi \to 1+a}\frac{\xi -1}{\xi}; $$ hence it suffices to prove for the map $\xi \to \frac{\xi -1}{\xi}$ with $a \neq -1$.

If $\xi \neq 0$, then $$ \bigg| \frac{\xi -1}{\xi} - \frac{a}{1+a} \bigg| = \frac{|\xi - (1+a)|}{|\xi||1+a|}; $$ if in addition $|\xi - (1+a)| < \frac{|1+a|}{2} =: \delta_{1}$, then by triangle inequality we have $||\xi| - 2\delta_{1}| \leq |\xi - (1+a)| < \delta_{1}$, implying that $\delta_{1} < |\xi|$, implying that $$ \frac{|\xi - (1+a)|}{|\xi||1+a|} < \frac{|\xi - (1+a)|}{2\delta_{1}^{2}} =: M_{\xi}; $$ given any $\varepsilon > 0$, we have $M_{\xi} < \varepsilon$ if in addition $|\xi - (1+a)| < 2\delta_{1}^{2}\varepsilon =: \delta_{2}$. All in all we conclude that, for every $\varepsilon > 0$, for $\xi \neq 0$ and $|\xi - (1+a)| < \min \{ \delta_{1}, \delta_{2} \}$ we have $$ \lim_{\xi \to 1+a}\frac{\xi-1}{\xi} = \frac{a}{1+a}. $$