proving convergence using delta-epsilon notation

Question

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!

Answer 1

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.

Answer 2

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 $|x-1|<\delta$?

Answer 3

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?

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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 :)