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I'm having difficulty proving whether the following statement is true or not:

For any function $f(x)$, if $$\lim_{x\to c} f(x) = 0$$, then $$\lim_{x\to c} {1\over f(x)} = ∞$$

I have tried making x a real number and tested different functions. I found that it is possible to solve, but I'm not too sure how to approach the "proving" part of the question.

Any help will be much appreciated!

Edit: I know this may sound rudimentary and rather stupid, but, is it sufficient as 'proof' to state something like:
since $$\lim_{x\to c} f(x) = 0$$, then $$\lim_{x\to c}{1\over f(x)} = {1\over 0}=∞$$ because $$lim_{x\to c}{1\over 0} $$ will always be infinity?

J.Chiang
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    Consider the function $f(x)=x$ and $c=0$ then $\lim_{x\to 0}\frac{1}{f(x)}=\lim_{x\to 0}\frac{1}{x}=DNE$ – Teh Rod Apr 02 '17 at 07:44
  • No. You can't do that. Remember the law on the limit of quotients? By the way, when you write $\infty$, do you mean it as $+\infty$? – Juniven Acapulco Apr 02 '17 at 08:14
  • How can I help if you dont answer my question? – Juniven Acapulco Apr 02 '17 at 08:15
  • Sorry for the late reply - and yes, I do mean it as +∞. I apologise for not understanding as it has been a few years since I last properly did maths and am only just revisiting everything. – J.Chiang Apr 02 '17 at 08:22

1 Answers1

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Hint:

$\lim\limits_{x\to h}\dfrac{f(x)}{g(x)}=\dfrac{\lim\limits_{x\to h} f(x)}{\lim\limits_{x\to h} g(x)}$

Also note that $\lim\limits_{x\to h} k=k$ for any constant $k$. Hope it helps.

Fawad
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