need confirmation of counterexample

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October 26th 2009, 06:31 PM
dannyboycurtis

need confirmation of counterexample

Hi! I have to find a counterexample to the following statement:
$\lim_{x\to \infty}f(\frac{1}{x})=L$ implies $\lim_{x\to 0}f(x)=L$

I assert that $f(x)=\left\{\begin{array}{lr}1:&x>0\\-1:&x\leq 0\end{array}\right\}$ is a viable counterexample.

Is this correct?
The limit at 0 does not exist, but the limit of f(x) as x goes to infinity will be 1.
Im just not 100% sure...
Any feedback would be appreciated.
October 26th 2009, 07:47 PM
tonio

Quote:

Originally Posted by dannyboycurtis

Hi! I have to find a counterexample to the following statement:
$\lim_{x\to \infty}f(\frac{1}{x})=L$ implies $\lim_{x\to 0}f(x)=L$

I assert that $f(x)=\left\{\begin{array}{lr}1:&x>0\\-1:&x\leq 0\end{array}\right\}$ is a viable counterexample.

Is this correct?
The limit at 0 does not exist, but the limit of f(x) as x goes to infinity will be 1.
Im just not 100% sure...
Any feedback would be appreciated.

Nice, simple and neat example...oh, and correct, too.

Tonio

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