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Math Help - there is a careless mistake somewhere whcih i cant spot in my equation

  1. #1
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    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!
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  2. #2
    Moo
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    Quote Originally Posted by alexandrabel90 View Post
    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
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  3. #3
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    Quote Originally Posted by alexandrabel90 View Post
    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!

    \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}= \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
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  4. #4
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    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.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    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.
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