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.
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
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}
$