# Math Help - Limits without use of L'Hospital and series expansion

1. ## Limits without use of L'Hospital and series expansion

Without using series expansion of $e^x$ and binomial expansion of $\sqrt{1+x^2}$ evaluate the limit :
$\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}$
Also,you cannot use the L'Hospital rule.

2. 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

3. Thanks Chris.I am myself on it.

4. Originally Posted by pankaj
Without using series expansion of $e^x$ and binomial expansion of $\sqrt{1+x^2}$ evaluate the limit :
$\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}$
Also,you cannot use the L'Hospital rule.
The rules do not prevent you from using a series expansion of:

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

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

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

(here $o(x^3)$ denotes a function such that the limit as $x$ goes to $0$ of this function divided by $x^3$ 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

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

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

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

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

Putting $x = 3t$

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

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

$
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}
$

$
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}
$

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

$
12L = 2
$

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

6. Originally Posted by pankaj

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

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

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

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

Putting $x = 3t$

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

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

$
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}
$

$
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}
$

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

$
12L = 2
$

$
L = \frac {1}{6}
$
You seem to be using:

$
\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

7. I thought it was a standard formula or a definition

8. Originally Posted by CaptainBlack
You seem to be using:

$
\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?

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

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

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

10. Well, there's a way to justify the value of that limit. Try to find it.

11. 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

12. Haha, yeah, sorry, I misexpressed myself. I actually mean, a substitution method that's all.