Thread: there is a careless mistake somewhere whcih i cant spot in my equation

find the limit of ( 9x +1 ) ^( cot x) as x tends to 0 from the positive direction.

to solve this question, i let y = ( 9x +1 ) ^( cot x)
hence ln y = (cot x) ln (9x+1)
as x tends to 0 from the positive direction, y tends to 1.

lim exp ( ln y) = exp ( lim (ln y)) = exp (0) = 1

but if i were to check back my answers, i did not get an answer of 1.

here is my working:

lim exp ( ln y) = exp ( lim (ln y)) = exp ( lim ( cot x ln ( 9x +1))) = exp (lim ( ln ( 9x +1 ) / tan x )) = exp ( 9 cos ^2 x / ( 9x+1)) by l hopital rule.
which will give me exp (9) instead of my answer of exp (o).

thanks!

Originally Posted by alexandrabel90

find the limit of ( 9x +1 ) ^( cot x) as x tends to 0 from the positive direction.

to solve this question, i let y = ( 9x +1 ) ^( cot x)
hence ln y = (cot x) ln (9x+1)
as x tends to 0 from the positive direction, y tends to 1.

lim exp ( ln y) = exp ( lim (ln y)) = exp (0) = 1

but if i were to check back my answers, i did not get an answer of 1.

here is my working:

lim exp ( ln y) = exp ( lim (ln y)) = exp ( lim ( cot x ln ( 9x +1))) = exp (lim ( ln ( 9x +1 ) / tan x )) = exp ( 9 cos ^2 x / ( 9x+1)) by l hopital rule.
which will give me exp (9) instead of my answer of exp (o).

thanks!

yep, where does the red sentence come from ? you didn't do anything in order to find 1 :O that's just a statement without justification

Originally Posted by alexandrabel90

find the limit of ( 9x +1 ) ^( cot x) as x tends to 0 from the positive direction.

to solve this question, i let y = ( 9x +1 ) ^( cot x)
hence ln y = (cot x) ln (9x+1)
as x tends to 0 from the positive direction, y tends to 1.


No, it doesn't....why do you think so??


lim exp ( ln y) = exp ( lim (ln y)) = exp (0) = 1

but if i were to check back my answers, i did not get an answer of 1.

here is my working:

lim exp ( ln y) = exp ( lim (ln y)) = exp ( lim ( cot x ln ( 9x +1))) = exp (lim ( ln ( 9x +1 ) / tan x )) = exp ( 9 cos ^2 x / ( 9x+1)) by l hopital rule.
which will give me exp (9) instead of my answer of exp (o).

thanks!

We use L'Hospital's rule in the 3rd equality, of course. Now end the exercise.

Tonio

this is how i interpreted it:

as x tends to 0, wont (cot x) ln (9x+1) tend to 0? this would mean that ln y will tend to 0 and hence y will tend to 1...

thank you for correcting my mistake.

Originally Posted by alexandrabel90

this is how i interpreted it:

as x tends to 0, wont (cot x) ln (9x+1) tend to 0?

Tonio's post shows exactly that this is not true.