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Math Help - L'Hopital's Rule...!

  1. #1
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    L'Hopital's Rule...!

    Determine if L'Hopital's rule applies in each case. Select all that apply.

    lim_y-->0 3^y/y^2

    lim_x-->pi (sin(3x))/(x-pi)

    lim_t-->0 (t^2+3t)/(cosh(t)-1)

    lim_z-->0 (e^(2z)-1)/e^z

    lim_theta-->0 (arctan(theta))/(3(theta))

    lim_x-->infinity (e^(-x))/(1+ln(x))

    lim_x-->0+ (cot(x))/(ln(x))

    lim_t-->infinity (ln(t))^2/t

    we haven't exactly gone over L'Hopital's rule yet, but im tryin to do it early, any help on how you can tell whether the rule will work or not, thanks...

    mathete
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  2. #2
    Junior Member teuthid's Avatar
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    l'hospital's rule:

    if:
    \lim f(x) =0\  and\  \lim g(x) =0

    or if:
    \left| \lim f(x) \right| =\infty\  and\  \left| \lim g(x) \right| =\infty

    then:
    \lim \frac{f(x)}{g(x)}=\lim \frac{f'(x)}{g'(x)}

    For your questions, just make sure that the limits of the quotients are of the forms \frac{0}{0} or \frac{\infty}{\infty} as described above.
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  3. #3
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    ok, so does that just mean that the if the limit of some function is 0 or infinity then l'hopital's rule applies?
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  4. #4
    Junior Member teuthid's Avatar
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    Not quite, it requires that the limit of the quotient of functions (you'll notice that all your examples are in or can be written in the form \frac{f(x)}{g(x)}) must be one of the indeterminate forms: \frac{0}{0} or \frac{\infty}{\infty} (0r -\frac{\infty}{\infty}). If the limit of the quotient were \frac{0}{\infty}, or \infty or 2 or anything else, then L'hospital does not apply.
    Last edited by teuthid; April 3rd 2008 at 10:31 AM. Reason: ammendment
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  5. #5
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    OK, do you think you could show me an example of one of the ones that i put up there...that would help, thanks
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  6. #6
    Junior Member teuthid's Avatar
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    maybe this will clear things up...



    \lim_{x\rightarrow 0} \left(\frac{t^2+3t}{\cosh(t)-1}\right)=\frac{\lim_{x\rightarrow 0} (t^2+3t)}{\lim_{x\rightarrow 0}(\cosh(t)-1)}=\frac{0}{0}\Rightarrow L'Hospital's law applies



    \lim_{x\rightarrow 0^{+}} \left(\frac{\cot(x)}{\ln(x)}\right)=\frac{\lim_{x\  rightarrow 0^{+}}(\cot x)}{\lim_{x\rightarrow 0^{+}}(\ln x)}=\frac{\infty}{-\infty}\Rightarrow L'Hospital's law applies



    \lim_{x\rightarrow \infty} \left(\frac{e^{-x}}{1+\ln x}\right)=\frac{\lim_{x\rightarrow \infty} e^{-x}}{\lim_{x\rightarrow \infty} 1+\ln x}=<br />
\frac{0}{\infty}\Rightarrow L'Hospital's law does not apply
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  7. #7
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    Hello, mathlete!

    ok, so does that just mean that the if the limit of some function
    is 0 or infinity then l'hopital's rule applies?
    I wouldn't put it quite like that . . .

    If the limit (as written) is an indeterminate form, then L'Hopital applies.


    There are several indeterminate forms.

    The two basic forms are: . \frac{0}{0}\text{ and }\frac{\infty}{\infty} . . . and L'Hopital can be applied.

    There are other indeterminate forms: . \infty - \infty,\;0^0,\;0^{\infty},\;\infty^0,\;1^{\infty}, \;\hdots
    Often these can be transformed into one of the two basic forms,
    . . then L'Hopital can be used.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    1)\;\;\lim_{y\to0}\frac{3^y}{y^2} \;=\;\frac{3^0}{0^2} \;=\;\frac{1}{0} . . . not indeterminate

    2)\;\;\lim_{x\to\pi}\frac{\sin(3x)}{x-\pi} \;=\;\frac{\sin(3\pi)}{\pi-\pi} \;=\;\frac{0}{0} . . . yes, indeterminate!

    3)\;\;\lim_{t\to0} \frac{t^2+3t}{\cosh t - 1} \;=\;\frac{0^2 + 3(0)}{\cosh(0) - 1} \;=\;\frac{0+0}{1-1} \;=\;\frac{0}{0} . . . yes, indeterminate!

    4)\;\;\lim_{z\to0}\frac{e^{2x}-1}{e^x} \;=\;\frac{e^0-1}{e^0} \;=\;\frac{1-1}{1} \;=\;\frac{0}{1} \;=\;0 . . . not indeterminate


    Get the idea?

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