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I want to proove that if $\lim_{x\to c}$f(x) =$m$ then $\lim_{x\to c} \frac{1}{f(x)}$ = $1/m$ where $m\not= 0$.

I tried solving it using the epsilon delta definition but was stuck and not able to move forward using the definition for all $\epsilon>0$ there is $\delta>0$ such that $|x-c|<\delta$ then $|f(x)-m|<\epsilon \Rightarrow m-\epsilon<f(x)<m+\epsilon$

After this step I am not able to move forward.

For $1/f(x)$ I need to find a $\epsilon'>0$ such that there is $\delta'>0$. How can I proceed?

Aditya Dwivedi
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Rathi_Ji
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    This might be helpful: https://math.stackexchange.com/questions/354632/show-that-fx-1-x-is-continuous-at-any-c-neq-0 – Jake28 Jun 05 '20 at 17:09

1 Answers1

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Let's assume, that $m>0$. So we can find $\delta_1$ neighbourhood of $c$ in which $\dfrac{m}{2}<|f(x)|$.

Let's consider $\epsilon > 0$.

In limit definition for $ \lim_{x \rightarrow c}f(x)=m$ we can take also any $\epsilon_1 > 0$. So let's take $\epsilon_1 = \dfrac{m^2}{2} \cdot \epsilon$. For such $\epsilon_1$ exist $\delta_2$ neighbourhood of $c$ in which $\left| f(x) - m \right| < \epsilon_1$

So in intersection of $\delta_1$ and $\delta_2$ neighbourhoods, will be $$ \left| \dfrac{1}{f(x)} - \dfrac{1}{m}\right| = \left| \dfrac{f(x)-m}{f(x)m} \right| < \dfrac{\epsilon_1}{|f(x)|m} < \dfrac{ 2}{m^2} \cdot \epsilon_1 = \epsilon $$

zkutch
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