Without using series expansion of $\displaystyle e^x$ and binomial expansion of $\displaystyle \sqrt{1+x^2}$ evaluate the limit :

$\displaystyle \lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}$

Also,you cannot use the L'Hospital rule.

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- Sep 25th 2008, 06:19 AMpankajLimits without use of L'Hospital and series expansion
Without using series expansion of $\displaystyle e^x$ and binomial expansion of $\displaystyle \sqrt{1+x^2}$ evaluate the limit :

$\displaystyle \lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}$

Also,you cannot use the L'Hospital rule. - Sep 26th 2008, 05:08 PMChris L T521
Its not that we don't want to solve it...its just that we can't think of a clever way to do it! We're discussing that same limit here.

--Chris - Sep 27th 2008, 06:55 AMpankaj
Thanks Chris.I am myself on it.

- Sep 27th 2008, 08:49 AMCaptainBlack
The rules do not prevent you from using a series expansion of:

$\displaystyle f(x)=e^x-x-\sqrt{1+x^2}$

Or avoiding series the fact that $\displaystyle f(0)=0,\ f'(0)=0,\ f''(0)=0 $ and the definition of (3rd) derivative to give:

$\displaystyle f(x)=\frac{x^3}{6}+o(x^3)$

(here $\displaystyle o(x^3)$ denotes a function such that the limit as $\displaystyle x$ goes to $\displaystyle 0$ of this function divided by $\displaystyle x^3$ goes to zero) which will allow the evaluation of the limit.

I suspect though that you might consider all this to be cheating as this is tantamount to either L'Hopital's rule and/or the use of series)

RonL - Oct 14th 2008, 04:26 PMpankaj
Please check my solution

The limit does not change if x is replaced by -x

i.e. $\displaystyle L=\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}$

$\displaystyle L=\lim_{x\to 0}\frac{-e^{-x}-x+\sqrt{1+x^2}}{x^3}$

$\displaystyle

L=\lim_{x\to 0}\frac{e^x-e^{-x}-2x}{2x^3}

$

Putting $\displaystyle x = 3t$

$\displaystyle L = \lim_{t\to 0}\frac {e^{3t} - e^{ - 3t} - 6 t }{54 t^3}$

$\displaystyle 54L = \lim_{t\to 0}\frac {(e^{t} - 1)^3 + (3e^{2t} - 3e^{t} + 1) - (e^{ - t} - 1)^3 - 3e^{ - 2t} + 3e^{ - t} - 1 - 6 t }{ t^3}$

$\displaystyle

54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + \frac {3e^{2t} - 3e^{ - 2t} - 3e^{t} + 3e^{ - t} - 6t}{t^3}

$

$\displaystyle

54L = \lim_{t\to 0}(\frac {e^t - 1}{t})^3 + (\frac {e^{ - t} - 1}{ - t})^3 + 48\frac {e^{2t} - e^{ - 2t} - 4t}{2(2t)^3} - 6\frac {e^t - e^{ - t} - 2t}{2(t)^3}

$

$\displaystyle 54L = 1 + 1 + 48L - 6L = 2 + 42L$

$\displaystyle

12L = 2

$

$\displaystyle

L = \frac {1}{6}

$ - Oct 14th 2008, 08:18 PMCaptainBlack
- Oct 15th 2008, 06:12 AMpankaj
I thought it was a standard formula or a definition

- Oct 15th 2008, 06:41 AMChop Suey
- Oct 15th 2008, 08:46 AMCaptainBlack
- Oct 15th 2008, 09:29 AMKrizalid
Well, there's a way to justify the value of that limit. Try to find it.

- Oct 15th 2008, 11:35 AMCaptainBlack
- Oct 15th 2008, 11:38 AMKrizalid
Haha, yeah, sorry, I misexpressed myself. I actually mean, a substitution method that's all.