# Math Help - limit with restricted ways of doing it

1. ## limit with restricted ways of doing it

Hey Gang:

Here is a limit I saw somewhere recently but can not remember where. I think on another site.

Anyway, the restrictions are no expansions on e, no binomial expansion on the radical term and no L'Hopital. Go figure.

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

I just thought it'd be cool to see some clever ways folks come up with.

2. Someone posted it here a few days ago but got no answers:

http://www.mathhelpforum.com/math-he...expansion.html

3. The solution is 1/6

4. Originally Posted by galactus
The solution is 1/6
Maple tells me that $\lim_{x\to 0}\frac{e^{x}-x-\sqrt{1-x^{2}}}{x^{3}}$ is undefined...

--Chris

5. Originally Posted by Chris L T521
Maple tells me that $\lim_{x\to 0}\frac{e^{x}-x-\sqrt{1-x^{2}}}{x^{3}}$ is undefined...

--Chris
It should be $\sqrt{1+x^2}$

EDIT: attached a graph

6. Yes, I just noticed that and changed my typo. Sorry about that.

7. It's been posted too in Mathlinks, but answers do include Taylor expansions content. (At least, we avoided L'Hôpital's Rule.)

Anyway, it's kinda hard, I can't see a solution yet.

May be someone else's gonna give a nice solution.

8. This same question was also posted here.

--Chris

9. Could the squeeze theorem be applied to this beast??? Or would it turn out to be very tedious?? If we do find a way to evaluate this with the given restrictions, it will be a tedious process any...unless someone is very clever...

--Chris

10. By taking derivatives we can prove on $(-1,1)$ we have:
$1+x+\frac{x^2}{2}+\frac{x^3}{6} \leq e^x \leq 1 + x + \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}$

(These inequalities are inspired by Talyor's theorem but they make no mention of the theorem).

Uses these inequalities we can squeeze the function. The limit is completed by rationalizing the numerator. Here is an illustration with the lower bound.

$\frac{\left( 1 + x+\frac{x^2}{2}+\frac{x^3}{6}\right) - x - \sqrt{x^2+1}}{x^3} \leq \frac{e^x - x - \sqrt{x^2+1}}{x^3}$

Now this is,
$\frac{\left( 1 + \frac{x^2}{2} + \frac{x^3}{6} \right) - \sqrt{x^2+1}}{x^3} \cdot \frac{\left( 1 + \frac{x^2}{2}+\frac{x^3}{6} \right) + \sqrt{x^2+1}}{\left( 1 + \frac{x^2}{2}+\frac{x^3}{6} \right) + \sqrt{x^2+1}} = \frac{\left( 1 + \frac{x^2}{2} + \frac{x^3}{6} \right)^2 - x^2 - 1 }{x^3 \left( 1 + \frac{x^2}{2} + \frac{x^3}{6} + \sqrt{x^2+1} \right)}$

When we expand the trinomial we get $1+\frac{x^2}{2}+\frac{x^2}{2} + \frac{x^3}{6} + \text{ higher degree terms}$

After canceling and dividing by $x^3$ and ingoring higher degree terms we get $\frac{1}{6}$ as a lower bound.

Now do the same thing with the second polynomial.

11. 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}
$

12. Originally Posted by pankaj
The limit does not change if x is replaced by -x
i.e. $L=\lim_{x\to 0}\frac{e^x-x-\sqrt{1+x^2}}{x^3}$
There is still a problem. You are assuming the limit exists.
You need to show the limit actually exists. (Perhaps you can use fact that this is a decreasing function somehow).

13. Who missed this one!

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

Let $\varphi=x^3$

Then this becomes

\begin{aligned}\lim_{\varphi\to{0}}\frac{e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}}{\varphi}&=\lim_{\varphi\to{0}}\frac{e^ {\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}-0}{\varphi-0}\\
&=\left(e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}\right)'\bigg|_{x=0}\\
&=\frac{1}{6}\end{aligned}

You did just say no L'hopitals or Series.

14. Originally Posted by Mathstud28
Who missed this one!

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

Let $\varphi=x^3$

Then this becomes

\begin{aligned}\lim_{\varphi\to{0}}\frac{e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}}{\varphi}&=\lim_{\varphi\to{0}}\frac{e^ {\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}-0}{\varphi-0}\\
&=\left(e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}\right)'\bigg|_{x=0}\\
&=\frac{1}{6}\end{aligned}

You did just say no L'hopitals or Series.

Hahaha...clever substitution!! How long did it take you to figure it out?

--Chris

15. Originally Posted by Chris L T521

Hahaha...clever substitution!! How long did it take you to figure it out?

--Chris
I dont know. Like a minute and a half?

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