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May 12th, 2016, 05:39 AM   #1
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Need help with proving conversion of a limit

Prove that


e^t-t-1
lim ------------------------------ = 1/2
t->0 t^2


without using L'Hospital's theorem.

Help! I tried anything i can think of!
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May 12th, 2016, 05:46 AM   #2
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So this is $\displaystyle \lim_{x\to 0}\frac{e^t- t- 1}{t^2}$? Obviously the simplest thing to do is to use L'Hopital's rule but if you don't want to, replace $\displaystyle e^t$ with its expression as a MacLaurin series: $\displaystyle \lim_{x\to 0}\frac{1+ t+ t^2/2+ t^3/6+ \cdot\cdot\cdot t- 1}{t^2}= \lim_{t\to 0}\frac{t^2/2+ t^3/6+ \cdot\cdot\cdot}{t^2}= \lim_{t\to 0}\frac{1}{2}+ \frac{t}{6}+ \cdot\cdot\cdot$.
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May 12th, 2016, 06:56 AM   #3
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How is differentiating simpler than replacing with the series expansion and cancelling terms?
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May 12th, 2016, 08:48 AM   #4
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I now have a solution without series.
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May 12th, 2016, 09:46 AM   #5
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Here we go.

Suppose that $\lim \limits_{t \to 0} {\mathrm e^t - t - 1 \over t^2} = L$. Then we write $t=-u$ to get $\lim \limits_{u \to 0} {\mathrm e^{-u} +u - 1 \over u^2} = L$, which is valid because we have a two-sided limit. We can write both of these in terms of different variables: $x$ say; and sum them.
$$\begin{aligned}2L &= \lim_{x \to 0} {\mathrm e^x - x - 1 \over x^2} + \lim \limits_{t \to 0} {\mathrm e^{-x} +x - 1 \over x^2} &
&= \lim_{x \to 0} \left({\mathrm e^x - x - 1 \over x^2} + {\mathrm e^{-x} +x - 1 \over x^2}\right) \\
&= \lim_{x \to 0} {\mathrm e^x -2 +\mathrm e^{-x} \over x^2} &
&= \lim_{x \to 0} {\left(\mathrm e^{\frac12x} - \mathrm e^{-\frac12x}\right)^2 \over x^2} \\
&= \left( \lim_{x \to 0} {\mathrm e^{\frac12x} - \mathrm e^{-\frac12x} \over x}\right)^2 &
&= \left( \lim_{x \to 0} {\left(\mathrm e^{\frac12x}-1\right) - \left(\mathrm e^{-\frac12x}-1\right) \over x}\right)^2 \\
&= \left( \lim_{x \to 0} {\mathrm e^{\frac12x}-1 \over x} - \lim_{x \to 0} {\mathrm e^{-\frac12x}-1 \over x}\right)^2 &
&= \left( \left. {\mathrm d \over \mathrm d x} \mathrm e^{\frac12x}\right|_{x=0} - \left. {\mathrm d \over \mathrm d x} \mathrm e^{-\frac12x}\right|_{x=0} \right)^2 \\
&= \left(\left.\frac12\mathrm e^{\frac12x}\right|_{x=0} + \left.\frac12\mathrm e^{-\frac12x}\right|_{x=0}\right)^2 &
&= \left(\frac12 + \frac12\right)^2 \\
&= 1 \\
L &= \frac12
\end{aligned}$$
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