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Math Help - proving differentiablity

  1. #1
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    proving differentiablity

    Show that if f:\mathbb{R}\rightarrow\mathbb{R} is differentiable at x=1,

    \lim_{x\to 1}{\frac{f(x^2)-f(1)}{x^2-1}}=f'(1)

    How do we go about proving it? Thanks in advance.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: proving differentiablity

    Hint By definition, f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1} .
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    Re: proving differentiablity

    Quote Originally Posted by FernandoRevilla View Post
    Hint By definition, f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1} .
    t\rightarrow{1}\Rightarrow{t^2}\rightarrow{1}, but t^2\rightarrow{1}\nRightarrow{t}\rightarrow{1}, so is it right to say
    f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}=\lim_{t^2\to 1}\frac{f(t^2)-f(1)}{t^2-1}=\lim_{t\to 1}\frac{f(t^2)-f(1)}{t^2-1} ?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: proving differentiablity

    Quote Originally Posted by alphabeta89 View Post
    t\rightarrow{1}\Rightarrow{t^2}\rightarrow{1}, but t^2\rightarrow{1}\nRightarrow{t}\rightarrow{1}, so is it right to say
    f'(1)=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}=\lim_{t^2\to 1}\frac{f(t^2)-f(1)}{t^2-1}=\lim_{t\to 1}\frac{f(t^2)-f(1)}{t^2-1} ?
    Use the substitution t=\sqrt{x} so, \lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\ldots
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    Re: proving differentiablity

    Quote Originally Posted by FernandoRevilla View Post
    Use the substitution t=\sqrt{x} so, \lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\ldots
    So is it correct to say that
    \lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\lim_{\sqrt{t}\to 1}\frac{f(t)-f(1)}{t-1}=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}=f'(1)?
    Last edited by alphabeta89; January 12th 2012 at 10:07 PM.
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