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Math Help - derivative question

  1. #1
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    derivative question

    (lim x -> 0) ln(1+x)/x=1

    Hi, I'm stuck on this question. The book wants me to prove it, using the definition of a derivative.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cityismine View Post
    (lim x -> 0) ln(1+x)/x=1

    Hi, I'm stuck on this question. The book wants me to prove it, using the definition of a derivative.
    Ok

    well we first note that \ln(1)=0

    So then knowing that we can legally subtract zero from an expression we see that

    \lim_{\varphi\to{0}}\frac{\ln(1+\varphi)}{\varphi}  =\lim_{\varphi\to{0}}\frac{\ln(1+\varphi)-\ln(1+0)}{\varphi-0}

    Now seeing that this matches up pretty well with

    f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}

    We can see that

    \lim_{\varphi\to{0}}\frac{\ln(1+\varphi)-\ln(1+0)}{\varphi-0}=\bigg(\ln(1+\varphi)\bigg)'\bigg|_{\varphi=0}=\  bigg(\frac{1}{\varphi+1}\bigg)\bigg|_{\varphi=0}=\  frac{1}{1+0}=1
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  3. #3
    Math Engineering Student
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    Well, just in case:


    \begin{aligned}<br />
   \underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}&=\underset{x\to \infty }{\mathop{\lim }}\,x\ln \left( 1+\frac{1}{x} \right) \\ <br />
 & =\underset{x\to \infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{x} \right)^{x} \\ <br />
 & =\ln \Bigg[ \underset{x\to \infty }{\mathop{\lim }}\,\left( 1+\frac{1}{x} \right)^{x} \Bigg] \\ <br />
 & =\ln e \\ <br />
 & =1. \\ <br />
\end{aligned}
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