$\epsilon$-$\delta$ proof that $\lim_{x \to 0} \frac{1}{1+x} =1.$ [duplicate]

I am trying to prove that:

$$\lim_{x \to 0} \frac{1}{1+x} =1.$$

Here's how I started my proof,

$$|f(x)−1| = \left|\frac{1}{1+x} -1 \right| = \frac{x}{1+x} < \frac{\delta}{1+x}$$

I don't know how to continue now.

If $|x|<\frac12$, then $1+x\in\left(\frac12,\frac32\right)$; in particular, $1+x>\frac12$, and therefore $\left|\frac x{1+x}\right|<2|x|$. So, take $\delta=\min\left\{\frac12,\frac\varepsilon2\right\}$, and then\begin{align}|x|<\delta&\implies\left|\frac x{1+x}\right|<2|x|\text{ (since $|x|<\frac12$)}\\&\implies\left|\frac x{1+x}\right|<\varepsilon\end{align}(since $|x|<\frac\varepsilon2$).