Evaluating a limit approaching infinity....?

To see what happens we need to break this down a bit.

For both pairs of parenthesis we have

2^x - 3^-x

3^-x, can be written as 3/x

What happens when x gets HUGE?

If x is HUGE, then that fraction approaches zero, so it can be left out, since it will have no mentionable effect on the equation.

So the lim of x --> positive infinity (2^x - 3^-x)/(2^x - 3^-x) =

2^x/2^x = 1

Again, notice that we can leave out 3^-x, cause for a large number, e.g. infinity, it has no mentionable effect on the equation, since the fraction is sooo super small compared to the 2^infinity.

Hope this helped

lim x--->positive infinity (2^x - 3^-x) / (2^x + 3^-x)

lim x--->positive infinity ( 2^x - 1 / [3^x] ) / ( 2^x + 1 / [ 3^x ] )

lim x--->positive infinity ( 2^x*3^x - 1 ) / ( 2^x*3^x + 1 )

lim x--->positive infinity ( 6^x - 1 ) / ( 6^x + 1 )

Form oo / oo .. so L'Hopital's rule:

lim x--->positive infinity ( 6^x*ln( 6 ) ) / ( 6^x*ln( 6 ) )

lim x--->positive infinity 1 = 1

~Or:

lim x--->positive infinity ( 2^x - 1 / [3^x] ) / ( 2^x + 1 / [ 3^x ] )

lim x--->positive infinity ( 2^x - 1 / [3^x] ) / lim x--->positive infinity ( 2^x + 1 / [ 3^x ] )

( lim x--->positive infinity 2^x - lim x--->positive infinity 1 / [3^x] ) / ( lim x--->positive infinity 2^x + lim x--->positive infinity 1 / [ 3^x ] )

( lim x--->positive infinity 2^x ) / ( lim x--->positive infinity 2^x )

lim x--->positive infinity ( 2^x ) / ( 2^x )

lim x--->positive infinity 1 = 1