just testing

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May 1st 2010, 04:02 AM
Moo

$x_n \stackrel{n \to\infty}{\longrightarrow} x_\infty$
May 1st 2010, 05:06 AM
Opalg

Quote:

Originally Posted by Moo

$x_n \stackrel{n \to\infty}{\longrightarrow} x_\infty$

One advantage of \overset over \stackrel is that there is a corresponding command \underset, so that you can write $x_n \underset{n \to\infty}{\longrightarrow} x_\infty$ . You can even combine the two, to get things like $x_n \underset{\text{Moo}}{\overset{n \to\infty}{\longrightarrow}} x_\infty$ (if ever you should wish to).
May 1st 2010, 05:58 AM
Moo

There is also $\text{LaTeX} \xrightarrow[Opalg] {age \to 100} \text{perfection}$
XD

I agree, I prefer overset :) I just wanted to show another way (until I remembered xrightarrow)
May 1st 2010, 07:42 AM
Opalg

Quote:

Originally Posted by Moo

There is also $\text{LaTeX} \xrightarrow[Opalg] {age \to 100} \text{perfection}$

Thanks (I think). (Dull)

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