lim (2^x - 3^-x)/2^x + 3^-x

x-> infinity

Thanks.

Do you mean
(2^x - 3^-x)/(2^x + 3^-x) or
[(2^x - 3^-x)/2^x] + 3^-x
?
in other words, is the second 3^-x in the denominator?

In any case, the limit of 3^-x as x-> infnity is zero. That should help you figure out the answer.

Yes it is in the denominator.
I just get infinity/infintiy, which is not an answer

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