Evaluate $\displaystyle \lim\_{x\to0^+} \left( \lim\_{y\to0^+} \left( x^y \right) \right)$
Evaluate $\displaystyle \lim\_{y\to0^+} \left( \lim\_{x\to0^+} \left( x^y \right) \right)$
$\displaystyle = \lim\_{x\to0^+} \left( x^0 \right) = \lim\_{x\to0^+} \left( 1 \right) = 1 \vphantom{\left(\lim\_{y\to0^+}\right)}$.
$\displaystyle = \lim\_{y\to0^+} \left( 0^y \right) = \lim\_{y\to0^+} \left( 0 \right) = 0 \vphantom{\left(\lim\_{x\to0^+}\right)}$.
So there is danger in exchanging limits.
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## Exchanging limits
#### Experiment
Evaluate $\displaystyle \lim\_{x\to0^+} \left( \lim\_{y\to0^+} \left( x^y \right) \right)$
Evaluate $\displaystyle \lim\_{y\to0^+} \left( \lim\_{x\to0^+} \left( x^y \right) \right)$
$\displaystyle = \lim\_{x\to0^+} \left( x^0 \right) = \lim\_{x\to0^+} \left( 1 \right) = 1 \vphantom{\left(\lim\_{y\to0^+}\right)}$.
$\displaystyle = \lim\_{y\to0^+} \left( 0^y \right) = \lim\_{y\to0^+} \left( 0 \right) = 0 \vphantom{\left(\lim\_{x\to0^+}\right)}$.
So there is danger in exchanging limits.
To resolve the question
> When can we exchange limits?
we need the concept of uniform continuity.