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Thread: Derivatives and limits

  1. #1
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    Question Derivatives and limits

    Hi,

    I hope someone can help. I want to know whether my thinking is right or not about the following question:

    Derivatives and limits-screen-shot-2017-09-06-8.32.22-pm.png


    For 15a - Not exactly sure how to interpreted this one and would really appreciate help on this.

    For 15 b - I interpreted this limit as the derivative for e^x at x = 2, which implies that the limit for 15b is equal to e^2.

    - Olivia
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  2. #2
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    Re: Derivatives and limits

    Oh I think I get 15a now! I believe that the limit represents the derivative for e^x at x = 0! The answer for that limit is 1 since the e^0 = 1. Yay
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  3. #3
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    Re: Derivatives and limits

    $\dfrac {d}{dx} e^x =\displaystyle \lim_{h\to 0} \dfrac{ e^{x+h} - e^x}{h}$

    $\left(\left . \dfrac{d}{dx} e^x \right |_{x=0} \right)= \displaystyle \lim_{h\to 0} \dfrac{ e^h -1}{h}$

    $\dfrac{d}{dx}e^x = e^x$

    $\left . e^x \right |_{x=0} = 1$

    $\displaystyle \lim_{h\to 0} \dfrac{ e^h -1}{h}=1$

    see if you can do the 2nd one.
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  4. #4
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    Re: Derivatives and limits

    Quote Originally Posted by otownsend View Post
    Oh I think I get 15a now! I believe that the limit represents the derivative for e^x at x = 0! The answer for that limit is 1 since the e^0 = 1. Yay
    ok.. let's look at 2nd one then :P

    same deal but $x=2$
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    Re: Derivatives and limits

    No.

    I think it's \left.\frac{\mathrm d}{\mathrm dx} e^{2+x}\right|_{x=0}

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  6. #6
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    Re: Derivatives and limits

    Yeah thanks! makes sense.
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  7. #7
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    Re: Derivatives and limits

    Your answer was perfectly good too, hence the
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  8. #8
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    Re: Derivatives and limits

    Quote Originally Posted by otownsend View Post
    Hi,

    I hope someone can help. I want to know whether my thinking is right or not about the following question:

    Click image for larger version. 

Name:	Screen Shot 2017-09-06 at 8.32.22 PM.png 
Views:	21 
Size:	134.0 KB 
ID:	38077


    For 15a - Not exactly sure how to interpreted this one and would really appreciate help on this.

    For 15 b - I interpreted this limit as the derivative for e^x at x = 2, which implies that the limit for 15b is equal to e^2.

    - Olivia
    Actually in its actual form, the solution does not need to get derivative of the given forms. What it all asks is to put the limits i.e. $h \to 0$. But if you put those limits in place, you will soon notice that the term will become $\frac{0}{0}$ which is an indeterminate form.
    Hence then you are bound to use the L'Hospital rule which means you have to derivate the Numerator and Denominator to get the proper answer. Hope this makes sense.
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  9. #9
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    Re: Derivatives and limits

    That's a circular argument. There's no need, no any real use for l'H˘pital here.
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  10. #10
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    Re: Derivatives and limits

    Quote Originally Posted by otownsend View Post
    Oh I think I get 15a now! I believe that the limit represents the derivative for e^x at x = 0! The answer for that limit is 1 since the e^0 = 1.
    That absolutely correct.

    Now for 15b, note that $\dfrac{e^{2+h}-e^2}{h}=e^2\cdot\dfrac{e^{h}-1}{h}$, hence we can use part a to answer part b.
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