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Thread: L Hopital Rule

  1. #1
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    L Hopital Rule

    Ok I have this

    as lim x --> 0 for

    $\displaystyle \frac{sin^2x}{x^2}$

    and $\displaystyle sin^2(0)=0$

    and $\displaystyle 0^2=0$

    so the derivative as lim x-->0

    $\displaystyle \frac{2sinxcosx}{2x}$

    and i plug in 0 i get 0 so the limit is 0?
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    Quote Originally Posted by sk8erboyla2004 View Post
    ...
    so the derivative as lim x-->0

    $\displaystyle \frac{2sinxcosx}{2x}$

    and i plug in 0 i get 0 so the limit is 0?
    No, you get $\displaystyle \frac00,$ which is an indeterminate form.

    For your limit, note that

    $\displaystyle \frac{\sin^2x}{x^2}=\left(\frac{\sin x}x\right)^2.$

    There is a special trigonometric limit that you can apply here.
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    Ok so where I can from here?
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    $\displaystyle \lim_{x\to0}\frac{\sin^2x}{x^2}=\lim_{x\to0}\left( \frac{\sin x}x\right)^2$

    $\displaystyle =\left(\lim_{x\to0}\frac{\sin x}x\right)^2$

    $\displaystyle =1^2=1$
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  5. #5
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    So do you only apply the L Hopital Rule When the lim of each denom and num are not both 0 or postive or negative infinity ?

    how about for this one ?

    lim x-->0

    $\displaystyle \frac{e^{-x}}{x}$

    I do not see how you got 1 it is my fault because we are taught to take the limit of each num and denom
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    Quote Originally Posted by sk8erboyla2004 View Post
    So do you only apply the L Hopital Rule When the lim of each denom and num are not both 0 or postive or negative infinity ?
    L'HŰpital's rule is not needed for this limit. You can use it (twice), but that is more work. You should have been taught the following two special trigonometric limits.

    $\displaystyle \lim_{x\to0}\frac{\sin x}x=1$

    $\displaystyle \lim_{x\to0}\frac{1-\cos x}x=0$

    how about for this one ?

    lim x-->0

    $\displaystyle \frac{e^{-x}}{x}$
    Rewrite, $\displaystyle \frac{e^{-x}}x=\frac1{xe^x}.$ Since $\displaystyle e^x$ is always positive, it should be clear that as $\displaystyle x\to0,\;\frac1{xe^x}$ tends to $\displaystyle \infty$ from the right, and $\displaystyle -\infty$ from the left. This limit does not exist.
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  7. #7
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    I was never taught the special trig limits for the first problem I did go on step further and got 1

    because i noticed L Hoptial rule can no longer be used until you reach a constant in either denom or numerator
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    Quote Originally Posted by sk8erboyla2004 View Post
    I was never taught the special trig limits for the first problem I did go on step further and got 1

    because i noticed L Hoptial rule can no longer be used until you reach a constant in either denom or numerator
    If you must use l'Hopital's Rule, you should note that

    $\displaystyle \frac{\sin^2 x}{x^2} \longrightarrow \frac{2 \sin x \cos x}{2x} = \frac{\sin (2x)}{2x} \longrightarrow \frac{2 \cos (2x)}{2} = \cos (2x)$
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  9. #9
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    I think I get the geist of the rule

    when ever you have f(x)/g(x) and both limits are 0/0, inf/inf, and -inf/-inf

    then you apply the rule
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