Given $a_n$ goes to $a>0$, show that $\frac{a_n}{1+a_n}$ goes to $\frac a{1+a}$

Question

Given $a_n$ goes to $a>0$, show that $\dfrac{a_n}{1+a_n}$ goes to $\dfrac a{1+a}$.

I know I have to use an epsilon-delta argument, but I am unsure as to how I can prove it. It makes sense in my mind as to why this is true, I just can't get it onto paper.

Answer 1

Hint: Observe \begin{align} \left|\frac{a_n}{1+a_n} - \frac{a}{1+a}\right| = \frac{|a_n-a|}{|(1+a)(1+a_n)|} \end{align}