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I want to solve this limit without the use of L'Hôpital's rule:

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

Alekos Robotis
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Matheus
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    Well...what's $e^x$? If it is defined to be $\sum \frac {x^n}{n!}$ then you can use that. – lulu Aug 06 '17 at 14:00
  • https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Aug 06 '17 at 14:06
  • Are allowed to use taylor's expansion? – haqnatural Aug 06 '17 at 14:10
  • I don't find the answers in the link all that great. The highest ranked answer doesn't even show the limits exist! If one allows the FTC, then simpler answers can be given. – zhw. Aug 06 '17 at 17:30

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I would Taylor Expand $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ and plug in to find $$\lim_{x\to 0}\frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x}{x^2}=\lim_{x\to 0}\frac{\sum_{n=2}^\infty \frac{x^n}{n!}}{x^2}=\lim_{x\to 0}\frac{\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots}{x^2}=\lim_{x\to 0} \frac{1}{2}+\frac{1}{6}x+\frac{1}{24}x^2+\cdots=\frac{1}{2}.$$

Alekos Robotis
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  • Well yeah, but L'Hopital is just using taylor expansions in disguise. – Alice Ryhl Aug 06 '17 at 15:00
  • @AliceRyhl. Not in general. ...l'Hopital's rule for finding $\lim f(x)/g(x)$ does not require $f'$' or $g''$ to exist, and even if $f$ or $g$ is infinitely differentiable they may not be analytic. – DanielWainfleet Aug 06 '17 at 16:17