Sunday, August 25, 2019

real analysis - How do i evaluate this limit :$lim_{x to 0} frac{e^x-x-1}{x²}$ without using Taylor expansion




I would like to evaluate this limit without using Taylor expansion:




$$\lim_{x \to 0} \frac{e^x-x-1}{x^2}$$ .



Note: by Taylor Expansion i have got :$\frac{1}{2}$ .



Thank u for any help .!!!!


Answer



METHODOLOGY $1$: Use L'Hospital's Rule Successively



Repeated use of L'Hospital's Rule reveals




$$\begin{align}
\lim_{x\to 0}\frac{e^x-x-1}{x^2}&=\lim_{x\to 0}\frac{e^x-1}{2x}\\\\
&=\lim_{x\to 0}\frac12 e^x=\frac12
\end{align}$$






METHODOLOGY $2$: Integral representation of the numerator




Note that we can write the numerator as



$$\begin{align}
e^x-x-1&=\int_0^x \int_0^t e^s \,ds\,dt\\\\
&=\int_0^x \int_s^x e^s\,dt\,ds\\\\
&=\int_0^x (x-s)e^s\,ds
\end{align}$$



Now, we can use the Mean-Value-Theorem for integrals to reveal




$$\begin{align}
e^x-x-1&=e^{s^*(x)}\int_0^x(x-s)\,ds\\\\
&=\frac12 x^2e^{s^*(x)}
\end{align}$$



for some value of $s^*(x) \in (0,x)$.



Finally, exploiting the continuity of the exponential function yield the coveted limit



$$\begin{align}

\lim_{x\to 0}\frac{e^x-x-1}{x^2}&=\lim_{x\to 0}\frac{\frac12 x^2e^{s^*(x)}}{x^2}\\\\
&=\frac12
\end{align}$$



as expected!


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