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) $ 

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
$$-\frac{x^3}2-ax^4\le n(x)\le-\frac{x^3}2+bx^4.$$
