How can I prove that $\lim _{x\to 0}\left(\frac{1}{f\left(x\right)}\right)=\frac{1}{L}$ using the $ε-δ $ argument

Question

I want to know how I can prove that $\lim _{x\to 0}\left(\frac{1}{f\left(x\right)}\right)=\frac{1}{L}$ using the $ε-δ $ argument?

Question: Suppose $f :\mathbb R \to\mathbb R$ is a function such that:

$\lim _{x\to 0}\left(f\left(x\right)\right)=L$

Prove that:

$\lim _{x\to 0}\left(\frac{1}{f\left(x\right)}\right)=\frac{1}{L}$ using the $ε-δ $ argument

My attempt:

I know that for every $ε>0$, there exist a $δ>0$ such that $|x-a| < δ$ implies that $|f(x) - L| < ε$

Am I allowed to do :

$|x-0| < δ$ implies $ |\frac{1}{f(x)} -$ $\frac{1}{L}| < ε$, so I need to prove this statement using the fact that $|f(x) - L| < ε$

Pick $ε = 1$, then $-1 + L < f(x) < 1 + L$ so then:

$$\frac{1}{L\:-1}>\:\frac{1}{f\left(x\right)}>\:\frac{1}{L+\:1}$$

Looking back:

$$ \left|\frac{1}{f(x)} -\frac{1}{L}\right| = \left|\frac{L - f(x) }{f(x)\times L} \right| < \frac{1}{L\:-1} \times \frac{1}{L} \times ε $$

And then I just don't know what to do, am I even on the right path?

Answer 1

You've already noted that $$\left|\frac{1}{f(x)} - \frac{1}{L}\right| = \frac{|L-f(x)|}{|f(x)||L|}.$$

The intuition is that when $f(x)$ is close to $L$, the numerator is close to zero, and the denominator is close to $L^2$, so the whole thing is also close to zero.

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Let's try to formalize this. Fix some $\epsilon > 0$.

Denominator

We can be a little crude with the denominator. The definition of $\lim_{x \to 0} f(x) = L$ implies that there exists $\delta_1$ such that $|f(x)-L| < L/2$ as long as $|x| < \delta$. Note that this implies $|f(x)| \ge L/2$ ("reverse triangle inequality") for $|x| < \delta_1$.

So, so far, we have $$\frac{|L-f(x)|}{|f(x)||L|} \le \frac{|L-f(x)|}{L^2/2}, \qquad \text{if $|x| < \delta_1$}.$$

Numerator

The numerator is easy to handle. The definition of $\lim_{x \to 0} f(x) = L$ implies that there exists $\delta_2$ such that $|f(x)-L| < 2\epsilon / L^2$ for all $|x| < \delta_2$.

So, combining with our previous work, we can consider $|x| < \min\{\delta_1, \delta_2\}$ (so that our previous work for $\delta_1$ still holds) and obtain $$\frac{|L-f(x)|}{|f(x)||L|} \le \frac{|L-f(x)|}{L^2/2} \le \frac{2 \epsilon L^2}{L^2/2} = \epsilon, \qquad \text{if $|x| < \min\{\delta_1, \delta_2\}$}$$

Answer 2

To make life easy, I'm going to assume $L$ is positive, although the same idea works for $L$ arbitrary. You are on the right track with writing

$$\left|\frac{1}{f(x)} - \frac{1}{L}\right| = \left|\frac{L-f(x)}{f(x)L}\right|=\frac{|L-f(x)|}{|f(x)L|}$$

You want this thing to get small. To do this, you need to make sure the numerator is small and the denominator is not too small.

Given $\epsilon > 0$, you want to find a $\delta > 0$ such that the above expression is less than $\epsilon$ whenever $x \in (-\delta,\delta)$.

Without loss of generality, you can always take $\epsilon < \frac{L}{2}$ (the difficulty is always for $\epsilon$ really small, not $\epsilon$ really big). Now, since $\lim\limits_{x\to 0} f(x) = L$, there exists a $\delta > 0$ such that $|L - f(x)| < \min\{\frac{2\epsilon}{L^2}, \epsilon\}$ whenever $x \in (-\delta, \delta)$.

For such $x$, this implies that $f(x) \in (L - \epsilon, L + \epsilon) \subset (L - \frac{L}{2}, L + \frac{L}{2})$, because of our assumption on $\epsilon$. In particular, $|f(x)| > \frac{L}{2}$, and hence $|f(x)L| > \frac{L^2}{2}$.

Thus

$$\left|\frac{1}{f(x)} - \frac{1}{L}\right| = \frac{|L - f(x)|}{|f(x)L|} < \frac{ (\frac{2\epsilon}{L^2})}{(\frac{L^2}{2})} = \epsilon.$$

If this proof seems a bit too complicated and unintuitive, it is this way because I have written it by combining the proofs of two general results into one: first, that a composition of two continuous functions is continuous, and second, that the function $\frac{1}{x}$ is continuous.

Answer 3

For the property to be true, you need $L\neq 0$. Your argument is basically correct, but I think it contains some small flaws. For example, if $L=1$, $1/(L-1)$ does not make sense.

What you can do is first choose a temporary $\delta_1>0$ that guarantees $$ |f(x)-L|<|L|/2 $$ which guarantees that the product $|f(x)L|$ is bounded below by $|L|^2/2$. then you choose a second $\delta=\delta_2$ such that (as you do) $$ \frac{|L-f(x)|}{|L|^2/2}<\varepsilon $$ Finally, you choose $\delta(\varepsilon)=\min\{\delta_1,\delta_2\}$.

Answer 4

Depending on how much "scaffolding" you want to have. If you just want a rigorous way of verifying your claim is correct with the $\epsilon-\delta$ definition of limit, you can use the limit property (which proofs can be easily found online, such as in Paul's Online Notes)

1. $$\lim_{x\to a}\left[ \frac{g(x)}{f(x)} \right] = \frac{lim_{x\to a}g(x)}{lim_{x\to a}f(x)}, \text{provided that } lim_{x\to a}f(x) \ne 0$$
2. $$\lim_{x \to a} c = c$$

But if you want an efficient proof directly from the definition, you can try the following (credit, still, to Paul's Online Notes "Proof of 4").

Because $\lim_{x \to 0}f(x) = L$, there is a $\delta_1 > 0$ such that

$$\begin{aligned} 0 < |x-0| < \delta_1 && \text{implies} && |f(x)-L| < \frac{|L|}{2} \end{aligned}$$

Then, $$\begin{aligned} |L| =& |L-f(x)+f(x)| && \text{just adding zero to }L \\ \le& |L-f(x)| + |f(x)| && \text{using the triangle inequality} \\ =& |f(x)-L|+|f(x)| && |L-f(x)|=|f(x)-L| \\ <& \frac{|L|}{2}+|f(x)| && \text{assuming that } 0 < |x-0| < \delta_1 \end{aligned}$$

Rearranging it gives $$\begin{aligned}|L| < \frac{|L|}{2}+|f(x)| &&\Rightarrow&& \frac{|L|}{2} < |f(x)| &&\Rightarrow &&\frac{1}{|f(x)|} < \frac{2}{|L|}\end{aligned}$$

Now, let $\epsilon > 0$. There is also a $\delta_2$ such that,

$$\begin{aligned} 0 < |x-0| < \delta_2 && \text{implies} && |f(x)-L| < \frac{|L|^2}{2} \epsilon \end{aligned}$$

Choose $\delta = \min{(\delta_1, \delta_2)}$. If $0<|x-0|<\delta$ we have,

$$\begin{aligned} \left|\frac{1}{f(x)} - \frac{1}{L}\right| =& \left| \frac{L-f(x)}{Lf(x)} \right| && \text{common denominators} \\ =& \frac{1}{\left|Lf(x)\right|} \left| L-f(x) \right| && \\ =& \frac{1}{|L|}\frac{1}{|f(x)|}|f(x)-L| && \\ <& \frac{1}{|L}\frac{2}{|L|}|f(x)-L| && \text{assuming that } 0<|x-0|<\delta \le \delta_1 \\ <& \frac{2}{|L|^2} \frac{|L|^2}{2} \epsilon && \text{assuming that } 0 < |x-0| < \delta \le \delta_2 \\ =& \epsilon && \end{aligned}$$