0

I know that the answer to the limit $$\lim_{x\to 0}\frac{e^x-x-1}{x^2}$$ is $1/2$ but I want to know how to solve it with my hand.

Qurultay
  • 5,224
Seokwon Lee
  • 1
  • 2
    https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Jun 14 '19 at 06:09
  • 1
    Use either series expansion or L'Hopital's Rule. – Kavi Rama Murthy Jun 14 '19 at 06:10
  • 1
    Use the taylor expansion: $e^x=\frac {x^0}{0!}+\frac {x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$ – For the love of maths Jun 14 '19 at 06:14
  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jun 14 '19 at 06:14
  • L'Hopital's Rule is easier. – alan23273850 Jun 14 '19 at 06:28
  • 1
    Without hands: you recognize that the negative terms cancel out the first two terms of the Taylor development of $e^x$, and the next coefficient is $\frac1{2!}$. –  Jun 14 '19 at 06:33

3 Answers3

3

Hint $$e^x-x-1=\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$ Using the Taylor series for $e^x$.

E.H.E
  • 23,280
2

L’Hospital’s Rule: When you have limit in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, differentiate the numerator and denominator and then apply the limit

For this problem, apply L’Hospital’s Rule twice.

Hari Prasad
  • 161
0

You can use series of $e^x$. A more quick way is using l-hopital rule

If putting the value in the limit gives $\frac{0}{0} or \frac{\infty}{\infty}$ then we can use this method. Simply differentiate the numerator and denominator until it is no more $\frac{0}{0} or \frac{\infty}{\infty}$ (as the case may be) form.

$\lim_{x\to 0}\frac{e^x-x-1}{x^2} = \lim_{x\to0} \frac{e^x-1}{2x}$ Again apply the rule.

$\lim_{x\to0} \frac{e^x}{2}=\boxed{\frac{1}{2}}$

Vineet
  • 846