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$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$

I give the two alternate ways so that no one gives them again in their answer:

Using L'Hospital: $$L=\lim_{x\to0}\frac{e^x-x-1}{3x^2}=\lim_{x\to0}\frac{e^x-1}{6x}=\lim_{x\to0}\frac{e^x}{6}=\frac16$$ Using Taylor: $$L=\lim_{x\to0}\frac{\color{red}{(1+x+x^2/2+x^3/6+O(x^4))}-1-x-x^2/2}{x^3}=\frac16$$ Using nothing(not infact): $$L=\lim_{x\to0}\frac{e^x-1-x^2/2}{x^3}\tag{sorry I omitted the x}\\ =\lim_{x\to0}\frac{e^{2x}-1-2x^2}{8x^3}\\ 8L=\lim_{x\to0}\frac{e^{2x}-1-2x^2}{x^3}\\ 7L=\lim_{x\to0}\frac{e^{2x}-e^x-3x^2/2}{x^3}\\ 7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{8x^3}\\ 56L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-6x^2}{x^3}\\ 49L=\lim_{x\to0}\frac{e^{4x}-2e^{2x}+e^x-9x^2/2}{x^3}=?$$

agha
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RE60K
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3 Answers3

2

$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$

If Limit exists: $$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}\\ L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{8x^3}\\ 8L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{x^3}\\ 7L=\lim_{x\to0}\frac{e^{2x}-e^x-x-3x^2/2}{x^3}\\ 7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-2x-6x^2}{8x^3}\\ 28L=\lim_{x\to0}\frac{\frac12e^{4x}-\frac12e^{2x}-x-3x^2}{x^3}\\ 21L=\lim_{x\to0}\frac{\frac12e^{4x}-\frac32e^{2x}+e^x-\frac32x^2}{x^3}\\ 21L=\lim_{x\to0}\frac{\frac12e^{8x}-\frac32e^{4x}+e^{2x}-6x^2}{8x^3}\\ 168L=\lim_{x\to0}\frac{\frac12e^{8x}-\frac32e^{4x}+e^{2x}-6x^2}{x^3}\\ 42L=\lim_{x\to0}\frac{\frac18e^{8x}-\frac38e^{4x}+\frac14e^{2x}-\frac32x^2}{x^3}\\ 21L=\lim_{x\to0}\frac{\frac18e^{8x}-\frac78e^{4x}+\frac74e^{2x}-e^x}{x^3}\\ 21L=\lim_{x\to0}\frac{\frac18e^x(e^x-1)^3(10e^x+6e^{2x}+3e^{3x}+e^{4x}+8)}{x^3}\\ 21L=\frac1828\\ \huge L=\frac16 $$

RE60K
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    Calculating a limit also involves showing that the limit exists. What you have proved above is that "if $L$ exists it must be $1/6$." The issue of existence can't be dealt with without further investigation which would require tools like taylor and LHR – Paramanand Singh Sep 22 '14 at 17:44
  • @ParamanandSingh I know the problem with this analysis, I have been told about it many times, putting that as anote from now on, see the edit – RE60K Sep 22 '14 at 17:45
  • One possible approach is to use the definition $e^{x} = \lim_{n \to \infty}(1 + (x/n))^{n}$. This gives the limit without much issue. – Paramanand Singh Sep 22 '14 at 17:46
  • @ParamanandSingh maybe you'll make an answer out of that? – RE60K Sep 22 '14 at 17:47
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I will deal with the case $x \to 0^{+}$ and leave the $x \to 0^{-}$ case for OP and other readers. Let us use the formula $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ We then have $$\begin{aligned}L &= \lim_{x \to 0^{+}}\dfrac{e^{x} - 1 - x - \dfrac{x^{2}}{2}}{x^{3}}\\ &= \lim_{x \to 0^{+}}\dfrac{\lim_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} - 1 - x - \dfrac{x^{2}}{2}}{x^{3}}\\ &= \lim_{x \to 0^{+}}\lim_{n \to \infty}\dfrac{\left(1 + \dfrac{x}{n}\right)^{n} - 1 - x - \dfrac{x^{2}}{2}}{x^{3}}\\ &= \lim_{x \to 0^{+}}\lim_{n \to \infty}\frac{1}{x^{3}}\left\{-\frac{x^{2}}{2n} + \dfrac{\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)}{3!}x^{3} + \cdots\right\}\\ &= \lim_{x \to 0^{+}}\lim_{n \to \infty}\left\{-\frac{1}{2nx} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)\left(1 - \dfrac{3}{n}\right)}{4!}x + \cdots\right\}\\ &= \lim_{x \to 0^{+}}\lim_{n \to \infty}f(x, n) - \frac{1}{2nx}\end{aligned}$$ where $f(x, n)$ is a finite sum with number of terms dependent on $n$. By using Monotone convergence theorem we can show that the sum $f(x, n)$ tends to a limit (as $n\to\infty$) dependent on $x$ say $f(x)$ and we can also show that $f(x) = (1/6) + o(x)$. The term $-1/2nx$ tends to $0$. It follows that $L = \lim_{x \to 0^{+}}f(x) = 1/6$.

Paramanand Singh
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Your answers are wrong because you omit the $-x\;$ term. And in your L'Hospital answer $e^x - x$ is not $0$ for $x=0!\,$ Your function goes to $+\infty$ for $x \rightarrow 0.$

gammatester
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