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 Post subject: Multiplicity of root
PostPosted: Mon Jan 11, 2016 1:54 pm 
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Given the function $f(x)=e^x-x-1$ prove that $0$ is a zero of $f$ of multiplicity $2$.

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 Post subject: Re: Multiplicity of root
PostPosted: Wed Aug 01, 2018 9:40 am 

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Tolaso J Kos wrote:
Given the function $f(x)=e^x-x-1$ prove that $0$ is a zero of $f$ of multiplicity $2$.


It suffices to prove that the limit $\displaystyle \lim \limits_{x \rightarrow 0} \frac{f(x)}{x^2}$ is finite. However,

\begin{align*}
\lim_{x\rightarrow 0} \frac{f(x)}{x^2} &= \lim_{x\rightarrow 0} \frac{e^x-x-1}{x^2} \\
&=\lim_{x\rightarrow 0} \frac{e^x-1}{2x} \\
&= \frac{1}{2}\lim_{x\rightarrow 0} e^x \\
&= \frac{1}{2}
\end{align*}

Done!

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