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

1. ## 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!

2. 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

3. 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!

$\displaystyle \lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\cos x}{\sin x}\,\ln(9x+1)=\lim_{x\to 0}\cos x\,\frac{\ln(9x+1)}{\sin x}=$ $\displaystyle \lim_{x\to 0}\cos x\cdot \lim_{x\to 0}\frac{9}{(9x+1)\cos x}=1\cdot \frac{9}{1\cdot 1} = 9$

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

Tonio

4. 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.

5. 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.