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 Post subject: Multiplicity of rootPosted: Mon Jan 11, 2016 1:54 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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 rootPosted: Wed Aug 01, 2018 9:40 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
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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