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Math Help - limit with l'hospitals rule

  1. #1
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    limit with l'hospitals rule

    Alright. So Ill show what Ive tried to do so far and hopefully someone can help me wrap it up. So using 'hospitals rule...

    lim sinxlnx
    x-> 0+

    so I took 2 approaches but came to a halt both ways

    FIRST WAY:

    lnx/(1/sinx) = FORM infin/infin

    so then limit equals

    = (1/x)/(-cotx cscx)

    dont know where to go from here

    SECOND WAY:

    sinx/(1/lnx) = FORM 0/0

    limit
    = cosx/lnx^2

    now im stuck again


    Yes, this post seems messy.
    So, how would I go about fully solving this problem
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrisc View Post
    Alright. So Ill show what Ive tried to do so far and hopefully someone can help me wrap it up. So using 'hospitals rule...

    lim sinxlnx
    x-> 0+

    so I took 2 approaches but came to a halt both ways

    FIRST WAY:

    lnx/(1/sinx) = FORM infin/infin

    so then limit equals

    = (1/x)/(-cotx cscx)

    dont know where to go from here

    SECOND WAY:

    sinx/(1/lnx) = FORM 0/0

    limit
    = cosx/lnx^2

    now im stuck again


    Yes, this post seems messy.
    So, how would I go about fully solving this problem
    the trick here is to realize that close to zero, \sin x \approx x. thus, close to zero, \sin x \ln x \approx x \ln x. if you don't like that, multiply by \frac xx to get x \frac {\sin x}x \ln x , taking the limit of the \frac {\sin x}x part, we arrive at the same thing.

    thus, \lim_{x \to 0^+} \sin x \ln x = \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac {\ln x}{\frac 1x} = - \lim_{x \to 0^+} \frac {- \ln x}{\frac 1x}

    now apply L'Hopital's
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  3. #3
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    I ended up with zero. How did I do?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrisc View Post
    I ended up with zero. How did I do?
    yes, that's correct

    here's a graph of the function, a second confirmation to your answer
    Attached Thumbnails Attached Thumbnails limit with l'hospitals rule-limit.jpg  
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  5. #5
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    Here is a justification what of what Jhevon did. He said x\approx \sin x so we can replace them. That is how they explain it in engineering classes we can do better and explain it like they do in math classes. What Jhevon is really saying is that \lim_{x\to 0^+}\sin x/x=1.
    That means,
    \lim_{x\to 0^+}\sin x\ln x = \lim_{x\to 0^+}\frac{\sin x}{x} \cdot x\ln x = \lim_{x\to 0^+}x\ln x.
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  6. #6
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    i guess ill throw this out as well
    its the last question Im having trouble with, and my brain is pretty much out of energy

    lim ((2x-3)/(2x+5))^(2x+1)
    x->infinity

    i took the ln of both sides, went to apply lhopitals rule, but no luck.
    ended up getting infinity on the top, but not the bottom
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrisc View Post
    i guess ill throw this out as well
    its the last question Im having trouble with, and my brain is pretty much out of energy

    lim ((2x-3)/(2x+5))^(2x+1)
    x->infinity

    i took the ln of both sides, went to apply lhopitals rule, but no luck.
    ended up getting infinity on the top, but not the bottom
    ok, let's do it your way. i still didn't end up with what you got:

    Let y = \lim_{x \to \infty} \left( \frac {2x - 3}{2x + 5} \right)^{2x + 1}

    \Rightarrow \ln y = \lim_{x \to \infty} (2x + 1) \ln \left( \frac {2x - 3}{2x + 5} \right)

    = \lim_{x \to \infty} \frac {\ln \left( \frac {2x - 3}{2x + 5}\right)}{\frac 1{2x + 1}} \to \frac 00

    Apply l'Hopital's

    \Rightarrow \ln y = \lim_{x \to \infty} \frac {\frac {16}{(2x + 5)^2}}{\frac {-2}{(2x + 1)^2}}

    = - \lim_{x \to \infty} \frac {8(2x + 1)^2}{2x + 5}

    = - \infty

    \Rightarrow y = e^{- \infty} = 0
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