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Thread: evaluating limits

  1. #1
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    evaluating limits

    evaluate:

    lim 3^(4+h) - 3^(4)
    h->0 h


    i think i remember something from calc that had to do with the definition of a derivative, and a variation of the formula looking something similiar to this... any help?? i dont know where to begin...

    thanks!
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  2. #2
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    Take note of $\displaystyle \frac{{3^{h + 4} - 3^4 }}{h} = 3^4 \frac{{3^h - 1}}{h}$.
    Now the problem reduces to finding if $\displaystyle f(x)=3^x$ then $\displaystyle f'(0)=?$
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  3. #3
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    okay, that does help a ton, and i know the answer is 3^4 * ln(3).

    but i know that from memorization, would you mind explaining how that is derived?
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  4. #4
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    $\displaystyle f(x) = 3^x \, \Rightarrow \,f'(0) = \lim _{h \to 0} \frac{{f(h) - f(0)}}
    {h}$
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  5. #5
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    Quote Originally Posted by Plato View Post
    $\displaystyle f(x) = 3^x \, \Rightarrow \,f'(0) = \lim _{h \to 0} \frac{{f(h) - f(0)}}
    {h}$
    how did i come up with ln3? i'm sorry but can you explain it in simpler terms i'm not as familiar with math as i used to be
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  6. #6
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    $\displaystyle \frac{d}{dx} a^x = a^x \ln{a}$
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  7. #7
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    Quote Originally Posted by sarabolha View Post
    evaluate:

    lim 3^(4+h) - 3^(4)
    h->0 h


    i think i remember something from calc that had to do with the definition of a derivative, and a variation of the formula looking something similiar to this... any help?? i dont know where to begin...

    thanks!
    finding the limit directly is a pain. unless you know how to find $\displaystyle \lim_{h \to 0} \frac {3^h - 1}h$ or equivalently, $\displaystyle \lim_{h \to 0} \frac {e^{h \ln 3} - 1}h$. (TPH or Kriz can probably help you with that )

    the rule is genrally derived by using the chain rule (once we know the derivative of $\displaystyle e^x$) and not by evaluating the limit directly.

    we use the fact that $\displaystyle e^{\ln X} = X$ and $\displaystyle \log_a x^n = n \log_a x$

    so, $\displaystyle \frac d{dx}a^x = \frac d{dx} e^{x \ln a} = \bigg( \frac d{dx} x \ln a \bigg) \cdot e^{x \ln a}$ by the chain rule. which simplifies to $\displaystyle \ln a \cdot a^x = a^x \ln a$ for $\displaystyle a > 0$

    so recognizing that it is the derivative limit and running the above proof is the way to go



    an alternative to find the limit without this, is to use a slightly modified version of the difference quotient, and some knowledge of hyperbolic functions.

    let $\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x - h)}{2h}$. this is equivalent to the previous definition. we just used the points $\displaystyle (x + h)$ and $\displaystyle (x - h)$ as opposed to $\displaystyle x$ and $\displaystyle (x + h)$.

    so, $\displaystyle \lim_{h \to 0} \frac {3^{4 + h} - 3^4}h = \lim_{h \to 0} \frac {3^{4 + h} - 3^{4 - h}}{2h}$

    $\displaystyle = 3^4 \cdot \lim_{h \to 0} \frac {3^h - 3^{-h}}{2h}$

    $\displaystyle = 3^4 \cdot \lim_{h \to 0} \frac {e^{h \ln 3} - e^{- h \ln 3}}{2h}$

    $\displaystyle = 3^4 \cdot \lim_{h \to 0} \frac {\sinh (h \ln 3)}h$

    $\displaystyle = 3^4 \cdot \ln 3 \cdot \lim_{h \to 0} \frac {\sinh (h \ln 3)}{h \ln 3}$

    $\displaystyle = 3^4 \cdot \ln 3 \cdot 1$

    $\displaystyle = 3^4 \cdot \ln 3$
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