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Math Help - Find the limit by evaluating the derivative only

  1. #1
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    Find the limit by evaluating the derivative only

    What does the title mean? What is it asking for? The question so far has consisted of:

    Find a function f and a number, (c), such that: \lim \limits_{h \to 0}\frac{(2+h)^5-32}{h}

    I found out that the answer was 80 but had to refresh my knowledge on how to expand binomials easily. I found this link that helped tremendously with that component.

    Now it's asking "Find the above limit by evaluating the derivative only". However this is a limit section of the textbook, we haven't learned the chain rule yet. Only the definition of the derivative, but it's not limiting my use of using the chain rule in this question. So what is it exactly asking and how would you approach this question? I would imagine what it is asking is easy but half the battle in math is understanding what it is asking for.
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    Re: Find the limit by evaluating the derivative only

    Quote Originally Posted by Maskawisewin View Post
    What does the title mean? What is it asking for? The question so far has consisted of:

    Find a function f and a number, (c), such that: \lim \limits_{h \to 0}\frac{(2+h)^5-32}{h}

    I found out that the answer was 80 but had to refresh my knowledge on how to expand binomials easily. I found this link that helped tremendously with that component.

    Now it's asking "Find the above limit by evaluating the derivative only". However this is a limit section of the textbook, we haven't learned the chain rule yet. Only the definition of the derivative, but it's not limiting my use of using the chain rule in this question. So what is it exactly asking and how would you approach this question? I would imagine what it is asking is easy but half the battle in math is understanding what it is asking for.
    What are you expecting this limit to equal?
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    Re: Find the limit by evaluating the derivative only

    Do you recognise that the "difference quotient" defining the derivative of [tex]x^5[/itex] at x= 2? What is the derivative of x^5? What is its value at x= 2?
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    Re: Find the limit by evaluating the derivative only

    Okay it took me a bit but I see now all it's asking is "go backwards". If this is the function plugged into the definition of the derivative then work backwards. It came across as very cryptic at first. Thank you.
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    Re: Find the limit by evaluating the derivative only

    Quote Originally Posted by Maskawisewin View Post
    Okay it took me a bit but I see now all it's asking is "go backwards". If this is the function plugged into the definition of the derivative then work backwards. It came across as very cryptic at first. Thank you.

    I have this exact same question and this doesn't really help me to understand what's being asked for here.
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    Re: Find the limit by evaluating the derivative only

    Quote Originally Posted by glenn22 View Post
    I have this exact same question and this doesn't really help me to understand what's being asked for here.
    note ...

    $\displaystyle f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

    for your specific problem, $f(x) = x^5$ and $a = 2$ ... the question wants you to determine the limit value which is $f'(2)$

    if $f(x) = x^5$ , then $f'(x) = 5x^4 \implies f'(2) = 5(2^4) = 80$
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    Re: Find the limit by evaluating the derivative only

    Quote Originally Posted by skeeter View Post
    note ...

    $\displaystyle f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

    for your specific problem, $f(x) = x^5$ and $a = 2$ ... the question wants you to determine the limit value which is $f'(2)$

    if $f(x) = x^5$ , then $f'(x) = 5x^4 \implies f'(2) = 5(2^4) = 80$
    I guess where this becomes confusing is that this work is all done before discussion about the power rule to find derivatives (as the OP mentioned), so am I then being asked to find the derivative the long way (ie. using the definition of the derivative) and then evaluate it at 2?
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    Re: Find the limit by evaluating the derivative only

    if using the definition is how you have to evaluate the derivative, then do it ...

    f(x) = x^5

    f(x+h) = (x+h)^5

    f'(x) = \lim_{h \to 0} \frac{(x+h)^5 - x^5}{h}

    f'(x) = \lim_{h \to 0} \frac{x^5+5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5 - x^5}{h}

    f'(x) = \lim_{h \to 0} \frac{5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5}{h}

    f'(x) = \lim_{h \to 0} \frac{h(5x^4+10x^3h+10x^2h^2+5xh^3+h^4)}{h}

    f'(x) = \lim_{h \to 0} 5x^4+10x^3h+10x^2h^2+5xh^3+h^4 = 5x^4

    now evaluate $f'(2)$
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    Re: Find the limit by evaluating the derivative only

    Quote Originally Posted by Maskawisewin View Post
    The question so far has consisted of:

    Find a function f and a number, (c), such that: \lim \limits_{h \to 0}\frac{(2+h)^5-32}{h}
    I think you mean: Find a function $f$ and a number $c$ such that

    \lim_{h\to0}\frac{(2+h)^5-32}h\ =\ c
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    Re: Find the limit by evaluating the derivative only

    Thanks for all your help guys. I believe i will answer the question by getting the derivative using the definition of the derivative (the long way) and then solving for 2, It may be more than what is required, but I don't think my teacher would dock me marks for doing extra work right?

    Thanks again everyone.
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