# Limits without use of L'Hospital and series expansion

• Sep 25th 2008, 06:19 AM
pankaj
Limits 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 PM
Chris L T521
Quote:

Originally Posted by pankaj
Nobody wants to solve this limit!

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 AM
pankaj
Thanks Chris.I am myself on it.
• Sep 27th 2008, 08:49 AM
CaptainBlack
Quote:

Originally Posted by pankaj
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.

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 PM
pankaj

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 PM
CaptainBlack
Quote:

Originally Posted by pankaj

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 $\displaystyle x = 3t$

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

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

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

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

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

$\displaystyle 12L = 2$

$\displaystyle L = \frac {1}{6}$

You seem to be using:

$\displaystyle \lim_{u \to 0} \frac{e^u-1}{u}=1$

How do you justify this without the use of L'Hopital or a series expansion?

CB
• Oct 15th 2008, 06:12 AM
pankaj
I thought it was a standard formula or a definition
• Oct 15th 2008, 06:41 AM
Chop Suey
Quote:

Originally Posted by CaptainBlack
You seem to be using:

$\displaystyle \lim_{u \to 0} \frac{e^u-1}{u}=1$

How do you justify this without the use of L'Hopital or a series expansion?

CB

Can't we use definition of the derivative?

$\displaystyle \lim_{u \to 0} \frac{e^u - e^0}{u-0} = (e^u)'~\text{at}~u = 0$
• Oct 15th 2008, 08:46 AM
CaptainBlack
Quote:

Originally Posted by Chop Suey
Can't we use definition of the derivative?

$\displaystyle \lim_{u \to 0} \frac{e^u - e^0}{u-0} = (e^u)'~\text{at}~u = 0$

Actually I think not, that is a 1 term series expansion, if yoy allow this you have to allow the method using the definition of the thrird derivative on the original problem.

CB
• Oct 15th 2008, 09:29 AM
Krizalid
Well, there's a way to justify the value of that limit. Try to find it.
• Oct 15th 2008, 11:35 AM
CaptainBlack
Quote:

Originally Posted by Krizalid
Well, there's a way to justify the value of that limit. Try to find it.

But we know what the limit is!

RonL
• Oct 15th 2008, 11:38 AM
Krizalid
Haha, yeah, sorry, I misexpressed myself. I actually mean, a substitution method that's all.