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

To solve the limit

lim (2^x - 3^-x)/(2^x + 3^-x) as x approaches infinity,

you can use the concept of leading terms and divide the numerator and denominator by the term with the highest exponent.

In this case, the term with the highest exponent is 2^x.

Dividing both the numerator and denominator by 2^x, we get:

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

Simplifying further:

lim ( 1 - (3^-x / 2^x) ) / ( 1 + (3^-x / 2^x) )

As x approaches infinity, the term (3^-x / 2^x) becomes negligibly small due to the negative exponent. Therefore, it tends to zero.

Thus, we can evaluate the limit as:

lim ( 1 - 0 ) / ( 1 + 0 ) = 1 / 1 = 1

Therefore, the limit of the function is equal to 1 as x approaches infinity.