# Thread: prove or disprove... involving sided limits

1. ## prove or disprove... involving sided limits

How can I prove the following:
$\displaystyle \lim_{x\to 0}f(x)=L$ implies $\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$
For that matter, how would I disprove this:
$\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$ implies $\displaystyle \lim_{x\to 0}f(x)=L$

2. Originally Posted by dannyboycurtis
How can I prove the following:
$\displaystyle \lim_{x\to 0}f(x)=L$ implies $\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$
For that matter, how would I disprove this:
$\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$ implies $\displaystyle \lim_{x\to 0}f(x)=L$
$\displaystyle \forall~\epsilon>0$, $\displaystyle \exists~\delta>0$ such that $\displaystyle |y-0|=|y|<\delta \implies |f(y)-L|<\epsilon$

Replace $\displaystyle y$ with $\displaystyle \frac{1}{x}$ and we have $\displaystyle |1/x|<\delta \implies |f(1/x)-L|<\epsilon$.

So $\displaystyle \forall~\epsilon>0$, $\displaystyle \exists~\delta>0$ such that $\displaystyle x>1/\delta \implies |f(1/x)-L|$.

This means $\displaystyle \lim_{x\to\infty}f(1/x)=L$.

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I cannot think of a counterexample for the second one. I'm sure someone else will, though.