Suppose f is real. Prove if L = lim f(x) (x goes to a) exists, then abs f(x) goes to abs L, as x goes to a.

Show that there is a funciton such that as x goes to a, abs f(x) goes to abs L, but the limit of f(x) does not exist.

Thanks.

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- Oct 6th 2007, 09:17 AMtaypezAdv Calc lim prf
Suppose f is real. Prove if L = lim f(x) (x goes to a) exists, then abs f(x) goes to abs L, as x goes to a.

Show that there is a funciton such that as x goes to a, abs f(x) goes to abs L, but the limit of f(x) does not exist.

Thanks. - Oct 6th 2007, 09:25 AMPlato
One of the most useful properties of absolute value is this:

$\displaystyle \left| {\left| x \right| - \left| y \right|} \right| \le \left| {x - y} \right|$.

It is also one of the least known/taught.

Using the $\displaystyle \varepsilon ,\delta$ definition for limit with the above, you problem ‘falls’ into your lap.