Find : $ \lim_{x\to 0} \frac{1+\sin x-\cos x+\log(1-x))}{x^3}$ 
Using squeeze theorem  find the following limit $$\lim_{x\to 0} \frac{1+\sin x-\cos x+\log(1-x))}{x^3}$$

My approach : 
I may only use the squeeze theorem. Please guide how to use this to solve the below question 
Suppose we have inequality : 
$h(x) \leq f(x) \leq g(x) $ 
$ \lim_{x\to c} h(x) \leq \lim_{x\to c} f(x) \leq \lim_{x\to c} g(x) $ 
$ \lim_{x\to c} h(x) \leq L \leq \lim_{x\to c} g(x) $ 
 A: Hint: Looking at taylor expansion of each functions, you can find upper and lower polynomial-type estimation for each functions on top.  For example 
$$  x - \frac{x^3}{6}  \leq \sin(x) \leq x - \frac{x^3}{6} +\frac{x^5}{5 !} $$ 
for all $x$ close to $0$.  You do this for each functions on top (at least up to the term $x^3$) then you can find squeezers. 
A: Hint: You can also apply the De l'Hopital's rule repeatedly:
$$
\lim_{x \rightarrow 0} \frac{1-\cos x+\sin x +\log (1-x)}{x^3}= \lim_{x \rightarrow 0} \frac{\cos x+\sin x -\frac{1}{1-x}}{3x^2}=...
$$
A: $sinx=x-{x^3}/{6}+o(x^3)$   
$cosx=1-x^2/2+o(x^3)$   
$log(1-x)=-x-x^2/2-x^3/3+o(x^3)$
Using this relationships you get that your limit is equal to $$lim_{x \rightarrow0} \frac{-x^3/2+o(x^3)}{x^3}$$
that goes to $-1/2$ as $x \rightarrow0 $
A: For small positive $x$,
$$0\le\cos x\le 1$$ and by definite integration in $[0,x]$,
$$0\le\sin x\le x.$$
Integrating again,
$$0\le 1-\cos x\le\frac{x^2}2$$
and
$$0\le x-\sin x\le\frac{x^3}6.$$
and
$$0\le\frac{x^2}2-1+\cos x\le\frac{x^4}{24}.$$
Similarly, from
$$6\le\frac6{(1-x)^4}\le12$$
integrating four times
$$-\frac{x^4}2\le\ln(1-x)+x+\frac{x^2}2+\frac{x^3}3\le-\frac{x^4}4.$$
By combining these results, we have for the numerator
$$-\frac{x^3}2-ax^4\le n(x)\le-\frac{x^3}2+bx^4.$$
