Prove using the epsilon-delta definition that $ \lim\limits_{x\to 0}\ \frac{1+x}{2-x}=\frac12$

Question

Prove using the epsilon-delta definition that $$\displaystyle \lim_{x\to 0}\ \dfrac{1+x}{2-x}=\frac12. $$ My try: $\displaystyle \lim_{x\to a}\ f(x)=b$ the $$|f(x)-b|<\epsilon$$ $$|f(x)-b|=|(1+x)/(2-x) -1/2)=|3x/(4-2x)|$$

but where do I go from here?

Answer 1

$$\forall x,\qquad |x|\leqslant\frac12\implies\left|\frac{1+x}{2-x}-\frac12\right|=\left|\frac{3x}{2(2-x)}\right|\leqslant|x|$$