how to prove that if given $\lim_{x\to a}f(x)=L$ then $\lim_{x\to a}\frac{1}{f(x)}=\frac{1}{L}$

Question

how to prove that if given $\lim_{x\to a}f(x)=L$ then $\lim_{x\to a}\frac{1}{f(x)}=\frac{1}{L}$

all i have so far is

$|\frac{f(x)-L}{Lf(x)}|<\epsilon$

And i don't know how to proceed from there

Link to other similar question

I know that the above link has the proof but I was confused why the Lf(x) was on the numerator not on the denominator at the end of the proof and was very confused

Answer 1

The assumption that $L\ne0$ is fundamental here, so I'll proceed under the condition $L\ne0$.

Here's the key: we need a lower bound on $|f(x)|$, which gives an upper bound for $1/|f(x)|$.

The inequality you have to solve is $$ \left|\frac{1}{f(x)}-\frac{1}{L}\right|<\varepsilon $$ Since $L\ne0$, there exists $\delta_0$ such that, for $0<|x-a|<\delta_0$, $$ |f(x)-L|<|L|/2 $$ which easily implies $$ |f(x)|>\frac{|L|}{2} $$ and so $$ \frac{1}{|f(x)|}<\frac{2}{|L|} $$ With the upper bond at our disposal, we can go on. Take $\delta>0$ such that $\delta\le\delta_0$ and, for every $x$ with $0<|x-a|<\delta$, $$ |f(x)-L|<\frac{L^2\varepsilon}{2} $$ Then $$ \left|\frac{1}{f(x)}-\frac{1}{L}\right|= \left|\frac{L-f(x)}{Lf(a+h)}\right|= \frac{1}{L}|f(x)-L|\frac{1}{|f(x)|}< \frac{1}{L}\frac{L^2\varepsilon}{2}\frac{2}{L} =\varepsilon $$

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But how did we get to $L^2\varepsilon/2$? It can be determined a posteriori. Let me rewrite:

Now take $\delta>0$ such that $\delta\le\delta_0$ and, for every $x$ with $0<|x-a|<\delta$, $$ |f(x)-L|<E $$ where $E$ will be determined later.

Then $$ \left|\frac{1}{f(x)}-\frac{1}{L}\right|= \left|\frac{L-f(x)}{Lf(a+h)}\right|= \frac{1}{L}|f(x)-L|\frac{1}{|f(x)|}< \frac{1}{L}E\frac{1}{L/2}=\frac{2E}{L^2} $$ and the last term equals $\varepsilon$ as soon as $$ E=\frac{L^2\varepsilon}{2} $$

Answer 2

We know that $\lim_{x\to a} C=C$, where $C$ is constant and moreover $\lim_{x \to a} \frac{f(x)}{g(x)}= \frac{\lim_{x\to a}f(x)}{\lim_{x \to a} g(x)}$. By using these two facts you can have your result immediately.

Answer 3

If $L\neq 0$, then given $\epsilon<|L|$ there is $\delta$ s.t. $|x-a|<\delta$ implies $|f-L|<\epsilon$. Hence $$\frac{-\epsilon }{L(L-\epsilon )}\leq \bigg|\frac{1}{f}- \frac{1}{L}\bigg| = \bigg|\frac{f-L}{fL}\bigg| \leq \frac{\epsilon }{L(L-\epsilon )} $$