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Thread: proving differentiablity

  1. #1
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    proving differentiablity

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

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

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

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

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

    Quote Originally Posted by FernandoRevilla View Post
    Use the substitution $\displaystyle t=\sqrt{x}$ so, $\displaystyle \lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\ldots$
    So is it correct to say that
    $\displaystyle \lim_{x\to 1}\frac{f(x^2)-f(1)}{x^2-1}=\lim_{\sqrt{t}\to 1}\frac{f(t)-f(1)}{t-1}=\lim_{t\to 1}\frac{f(t)-f(1)}{t-1}=f'(1)$?
    Last edited by alphabeta89; Jan 12th 2012 at 09:07 PM.
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