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!
 A: Hint:
Consider:
$$|xf(x)-1| = |xf(x)-x+x-1| \leq |xf(x)-x| + |x-1|$$
I've just used the triangle inequality here. So:
$$|xf(x)-1| \leq |x| \cdot |f(x)-1| + |x-1|$$
Now, to make life just a little bit easy for yourself, notice that:
$$|x| = |x-1+1| \leq |x-1| + 1$$
Hence, all in all, we can write:
$$|xf(x)-1| \leq |x-1| \cdot |f(x)-1| + |f(x)-1| + |x-1|$$
Now, can you use this as well as the facts you've been given to prove the result?

The next part will be the solution in full, for your reference. It's for reference only. You should only look at it when you've given this your best shot.

 Let $\epsilon > 0$ be given. We want to show that:
 $$\exists \delta > 0: \forall x \in \mathbb{R} \setminus \{1\}: |x-1| < \delta \implies |xf(x)-1| < \epsilon$$
 First of all, you know that there exists a $\delta_1 > 0$:
 $$|x-1| < \delta_1 \implies |f(x)-1| < \frac{1}{3}\epsilon$$
 You know this because $f(x) \to 1$. Let's look at the estimate  >! I derived in the hint above:
$$|xf(x)-1| \leq |x-1| \cdot |f(x)-1| + |f(x)-1| + |x-1|$$
 It would actually be really, really nice if I could just
 replace $|x-1| \cdot |f(x)-1|$ with
$1 \cdot \frac{1}{3}\epsilon$ and $|f(x)-1|$ with $\frac{1}{3}\epsilon$ and
  $|x-1|$ with $\frac{1}{3} \epsilon$. So, the idea here is
   this. Let's pick a $\delta > 0$ such that
 $\delta = \min\{1,\delta_1,\frac{1}{3}\epsilon\}$. What this means is >! that:
 $$\delta \leq 1 \ , \delta \leq \delta_1 , \ \delta \leq \frac{1}{3}\epsilon$$
 So, now, let's assume that $|x-1| < \delta$. That means that  >! $|x-1| < 1$ and $|x-1| < \delta_1$ and $|x-1| < \epsilon$. So, >! I can conveniently pick and choose where I want to use these  >! inequalities in my final estimate. In particular:
 $$|xf(x)-1| \leq |x-1| \cdot |f(x)-1| + |f(x)-1| + |x-1| < 1 \cdot \frac{1}{3} \epsilon + \frac{1}{3}\epsilon + \frac{1}{3}\epsilon = \epsilon$$
 But that proves what we wanted so we're done :)

A: 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 $|x-1|<\delta$?
A: The details of the $\delta-\varepsilon$ become obvious if you center your quantities around $1$, rather than akwardly adding and subtracting quantities:
Write $x = 1 + h$, $f(1 + h) = 1 + r(h)$. We have
$$xf(x) = (1 + h)(1 + r(h)) = 1 + r(h) + h + hr(h).$$
From here, given $\varepsilon \in (0, 1]$, we need to find $\delta > 0$ such that $|r(h) + h + hr(h)| \leq \varepsilon$ for all $0 < |h| \leq \delta$. But this is easy since, by assumption, we can find such $\delta$ for $r(h)$, and obviously we can find such $\delta$ for $h$. Taking the minimum of these two deltas and using the triangle inequality, we get
$$|r(h) + h + hr(h)| \leq 2\varepsilon + \varepsilon^2 \leq 3\varepsilon.$$
The factor of $3$ can be removed by replacing $\varepsilon$ with $\varepsilon/3$, so the proof is complete.
