<s>[INST] Find the limit $\lim_{x\to 0^-}| \left( 1+x^{3} \right)^{1/2}-1-x^{5} |/(\sin x-x)$ I am studying for my first calculus exam (well, it's half an exam), and of course we have to solve limits, without using L'Hospital rule, and using asymptotic analysis.
I can't solve this one $$\lim_{x\rightarrow 0^{-}}\frac{\left| \left( 1+x^{3} \right)^{\frac{1}{2}}-1-x^{5} \right|}{\sin x-x}$$
Can you guide me through the solution? Thanks a lot. 
 [/INST] Look at
$f(x)
=\frac{ ( 1+x^{3} )^{\frac{1}{2}}-1-x^{5} }{\sin x-x}
$.
By "asymptotic analysis",
as $x \to 0$,
$(1+x^3)^{\frac12}
\approx 1+\frac12 x^3 + \frac{\frac12 (-\frac12)}{2}(x^3)^2
= 1+\frac12 x^3 - \frac18 x^6
$
and
$\sin(x)
\approx x-\frac{x^3}{6}+\frac{x^5}{120}
$
so
$\sin(x)-x
\approx -\frac{x^3}{6}+\frac{x^5}{120}
$.
Therefore,
using the fact that
$ax^n+bx^m
\approx ax^n
$
as $x \to 0$
if $1 \le n < m$
and
$a\ne 0$,
$f(x)
\approx \frac{(1+\frac12 x^3- \frac18 x^6)-1-x^5}{-\frac{x^3}{6}+\frac{x^5}{120}} =\frac{\frac12 x^3 - \frac18 x^6-x^5}{-\frac{x^3}{6}+\frac{x^5}{120}}
\approx \frac{\frac12 x^3}{-\frac{x^3}{6}}
=-3
$.
</s>