Question

PLEASE, HELP!!!!
lim =(9^x-5^x)/x
x→0

Answer 1

do you know l'hospital?

Answer 2

It can be written as
lim (9^x/x) - lim (5^x/x)
x->0 x->0

Answer 3

Now use L'hospital rule

Answer 4

you need to keep the original form to have a 0/0 .

Answer 5

So it becomes
lim ln9*9^x/1 - lim ln5*5^x/1
x->0 x->0

Answer 6

So it is ln9-ln5 right?

Answer 7

yes, but you really are not allowed to apply l'hospital's rule to expressions like 1/0 (which is what you did).
to be entirely correct, work with the original expression... it gives ln(9)- ln(5) = ln(9/5)

Answer 8

oh.

Answer 9

so , do u know the formula
\[\lim_{x \rightarrow 0}\frac{ a^x-1 }{ x }=??\]

Answer 10

ok, it would be ln a
write your numerator as \((9^x-1)-(5^x-1)\)
then separate out into 2 limits as
\[\lim_{x \rightarrow 0}\frac{ 9^x-1 }{ x }\quad-\lim_{x \rightarrow 0}\frac{ 5^x-1 }{ x }\]
so u get ln 9 - ln 5 = ln (9/5).
got this?