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Math Help - [SOLVED] Prove two definitions of derivative is equivalent

  1. #1
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    [SOLVED] Prove two definitions of derivative is equivalent

    Hi, everyone! I have a minor problem proving the two following definitions of the derivative are equivalent:

    f is differentiable iff \lim_{x->x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}} exists
    and
    f is differentiable iff \lim_{t->0}\frac{f(x_{0}+t)-f(x_{0})}{t} exists.
    I know that i can let t=x - x_{0}, so that x=x_{0}+t and \frac{f(x) - f(x_{0})}{x - x_{0}}=\frac{f(x_{0} + t) - f(x_{0})}{t}, but then how can i explain the limits part, when in first case i am taking limit as x approaches x_{0} and in second as t approaches 0?

    Thanks in advance!
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  2. #2
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    First, be aware that I not really sure that I understand you difficulty here.
    That said, surely it is clear to you that this is true:
    \lim _{x \to x_0 } \left( {x - x_0 } \right) = 0\;\& \,\lim _{t \to 0} t = 0.

    Now tell us what you don't understand.
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  3. #3
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    Yea, i'm aware of that, but i'm not sure how I can apply that knowledge to my problem, since as far as I am concerned i'm not taking limit of t as t goes to 0, but rather limit of difference quotient as t goes to 0, where difference quotient need not be t. I guess what I really wanted to know is whether \lim_{x-x_0\to 0}(Y)=\lim_{x \to x_0}(Y) is generally true for any Y. Correct me if i'm totally off the track.
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  4. #4
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    Quote Originally Posted by vanishingpoint View Post
    Hi, everyone! I have a minor problem proving the two following definitions of the derivative are equivalent:

    f is differentiable iff \lim_{x->x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}} exists
    and
    f is differentiable iff \lim_{t->0}\frac{f(x_{0}+t)-f(x_{0})}{t} exists.
    I know that i can let t=x - x_{0}, so that x=x_{0}+t and \frac{f(x) - f(x_{0})}{x - x_{0}}=\frac{f(x_{0} + t) - f(x_{0})}{t}, but then how can i explain the limits part, when in first case i am taking limit as x approaches x_{0} and in second as t approaches 0?

    Thanks in advance!

    Since you defined t=x-x_0 , it follows at once that if x\rightarrow x_0 , then t\rightarrow 0, and the other way around. This is EXACTLy the reason why both limits above are equivalent.

    Tonio
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  5. #5
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    Thanks a lot, Tonio!
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