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Math Help - Don't understand some part of l'hopital's rule

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    Don't understand some part of l'hopital's rule

    6. c) Use l'Hopital's rule to evaluate lim x->0 e^x - e^-x -2x / x - sin x

    step 1 = e^x + e^-x -2 / 1 - cos x

    step 2 = e^x - e^-x / sin x

    step 3 = e^x + e^-x / cos x

    step 4 = above is = to 2/1

    I do not understand how 1-cosx becomes sin x, I thought it would be -sinx. Secondly, I do not understand how step 3 is equal to 2/1.
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    Quote Originally Posted by thekrown View Post
    6. c) Use l'Hopital's rule to evaluate lim x->0 e^x - e^-x -2x / x - sin x

    step 1 = e^x + e^-x -2 / 1 - cos x

    step 2 = e^x - e^-x / sin x

    step 3 = e^x + e^-x / cos x

    step 4 = above is = to 2/1

    I do not understand how 1-cosx becomes sin x, I thought it would be -sinx. Secondly, I do not understand how step 3 is equal to 2/1.
    Please either use LaTeX or correct use of brackets. What you have posted is very difficult to read.

    If I am reading it correctly, you are asked to evaluate

    \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}}

    using L'Hospital's Rule...


    First check that it is indeterminate.

    You should see that it tends to

    \frac{e^0 - e^0 - 2(0)}{0 - \sin{0}}

     = \frac{0}{0}.


    So you can use L'Hospital.

    Now, take the derivative of the top, and the derivative of the bottom.

    You will have

    \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x} - 2}{1 - \cos{x}}.


    Now see what happens when you let it \to 0.

    You should get

    \frac{e^0 + e^0 - 2}{1 - \cos{0}}

     = \frac{0}{0}.


    It is still indeterminate, so use L'Hospital again.

    \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x} - 2}{1 - \cos{x}} = \lim_{x \to 0}\frac{e^x - e^{-x}}{\sin{x}}

    (note that since the derivative of \cos{x} = -\sin{x} then the derivative of -\cos{x} = \sin{x}.)

     = \frac{e^0 - e^0}{\sin{0}}

     = \frac{0}{0}.


    Use L'Hospital again...

    \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x} - 2}{1 - \cos{x}} = \lim_{x \to 0}\frac{e^x - e^{-x}}{\sin{x}} = \lim_{x \to 0}\frac{e^x + e^{-x}}{\cos{x}}

     = \frac{e^0 + e^0}{\cos{0}}

     = \frac{1 + 1}{1}

     = 2.
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