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Math Help - Finding a limit?

  1. #1
    Newbie RangerKimmy's Avatar
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    Finding a limit?

    The problem:

    Find the limit as h→0 for (e^(2+h)-e^2) / h

    I really have no idea how to do this.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by RangerKimmy View Post
    The problem:

    Find the limit as h→0 for (e^(2+h)-e^2) / h

    I really have no idea how to do this.
    Consider this:
    \lim_{h\to 0}\frac{e^{2+h}-e^2}{h}=\lim_{h\to 0}\frac{e^2(e^h-1)}{h}=e^2\cdot\lim_{h\to0}\frac{e^h-1}{h}

    Have you seen the remaining limit \lim_{h\to0}\frac{e^h-1}{h}=\lim_{h\to0}\frac{e^h-e^0}{h} before?
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by RangerKimmy View Post
    The problem:

    Find the limit as h→0 for (e^(2+h)-e^2) / h

    I really have no idea how to do this.
    As h->0, you get \frac{0}{0}

    Use L'Hopital's rule

    The derivative of the numerator is:

    \frac{d}{dh} (e^{2+h} -e^2) = \frac{d}{dh} (e^2 \times e^h - e^2) = e^2.e^h-0=e^{2+h}

    and the derivative of the denominator is \frac{d}{dh} h = 1

    by L'hopital's rule , you then have:

    \lim_{h->0} \frac{e^{2+h}}{1}
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  4. #4
    Newbie RangerKimmy's Avatar
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    Quote Originally Posted by Failure View Post
    Consider this:
    \lim_{h\to 0}\frac{e^{2+h}-e^2}{h}=\lim_{h\to 0}\frac{e^2(e^h-1)}{h}=e^2\cdot\lim_{h\to0}\frac{e^h-1}{h}

    Have you seen the remaining limit \lim_{h\to0}\frac{e^h-1}{h}=\lim_{h\to0}\frac{e^h-e^0}{h} before?
    I follow what you have done.
    However, if that limit is something I should already know, then no, I haven't seen it before. Does it equal 1?
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by RangerKimmy View Post
    I follow what you have done.
    However, if that limit is something I should already know, then no, I haven't seen it before. Does it equal 1?
    Well it's the slope of the tangent of the curve y=e^x at the point (0,1), that is: the limit of the slope of the secant with one fixed point in (0,e^0)=(0,1) and the other point moving towards that fixed point ( h\to 0).
    Yes, this limit has the value 1.
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  6. #6
    Newbie RangerKimmy's Avatar
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    Quote Originally Posted by Failure View Post
    Well it's the slope of the tangent of the curve y=e^x at the point (0,1), that is: the limit of the slope of the secant with one fixed point in (0,e^0)=(0,1) and the other point moving towards that fixed point ( h\to 0).
    Yes, this limit has the value 1.
    Ohhh. Okay, thanks! This has helped tremendously!
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