proving convergence using delta-epsilon notation 
Let $f:\mathbb{R} \backslash\{ 1 \} \rightarrow \mathbb{R}$ be a function such that $f(x)\rightarrow 1$ as $x \rightarrow 1$. Using $\epsilon - \delta$ and without using any Limit Rules prove that $xf(x) \rightarrow 1$ as $x \rightarrow 1$.

My approach is, for $\delta\ > 0,$ there exist $\epsilon>0$ st $\lvert x-1\rvert <\delta,$ $\lvert f(x)-1\rvert< \epsilon$.
So
$$\lvert xf(x) -1\rvert<\lvert(\delta+1)f(x)-1\rvert<\lvert2f(x)-1\rvert$$
for $\delta <1$.
Please give some hints on how to approach this, thanks!

Hint: I would write:
$$
|xf(x)-1|=|f(x)-1+f(x)x-f(x)|\leq|f(x)-1|+|f(x)||x-1|
$$
For given $\epsilon>0$, let $\delta_0>0$ be small enough so that whenever $|x-1|<\delta_0$, we have:
\begin{aligned}
|f(x)-1|&<\epsilon/2,\\
|f(x)-1|&<1\implies|f(x)|<2.
\end{aligned}
Now, with $\delta\equiv\min\{\delta_0,\epsilon/4\}$, what happens whenever $|x-1|<\delta$?
