Prove that if $f(x)=L$ for deleted neighbourhood of $a$ then $\frac1{f(x)}=\frac1L$

Question

If I have the function $f:V\setminus \{a\}\to V'$ and $\lim_{x\to a}f(x)=L\ne0$, can I show that $\lim_{x\to a}\frac1{f(x)}=\frac1L$ using the $\varepsilon-\delta$ definition?

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What I tried

$$0<|x-a|<\delta\implies \left|\frac1{f(x)}-\frac1L\right|<\varepsilon$$I tried to get from $\left|\frac1{f(x)}-\frac1L\right|$ to $|f(x)-L|$ somehow: $$\left|\frac1{f(x)}-\frac1L\right|<\varepsilon\implies-\varepsilon<\frac1{f(x)}-\frac1L<\varepsilon\implies \frac{f(x)L}{L-f(x)}>\frac1\varepsilon\implies\frac{L-f(x)}{f(x)L}<\varepsilon\\\implies \frac{|L-f(x)|}{|f(x)L|}<\varepsilon\implies|f(x)-L|<|f(x)L|\varepsilon$$The problem is that I don't know how can I get rid of the $|f(x)L|$.

Answer 1

From the existence of limit we have that $\exists \delta_1 >0$ s.t. $0<|x-a|<\delta_1 \implies |f(x) - L| < 1$, so we have that $|f(x)| < 1 + |L|$ by the Triangle Inequality. Now also $\exists \delta_2 >0$ s.t. $0<|x-a|<\delta_2 \implies |f(x) - L| < \frac{\epsilon}{|L|(|L|+1)}$.

Then choose $\delta = \min(\delta_1,\delta_2)$ and we have:

$$0 < |x - a| < \delta \implies \left | \frac 1{f(x)} - \frac 1L\right| < |f(x)||L|\frac{\epsilon}{|L|(|L| + 1)} < \epsilon $$

Hence the proof.