<s>[INST] Evaluate a limit using Taylor series Let $$\lim\limits_{x\to 0}\frac{({\ln(1+x) -x +\frac{x^2}{2})^4}}{(\cos(x)-1+\frac{x^2}{2})^3}$$
Now, I know that I should utilize Taylor polynomial.
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...$
$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ...$
Plugin it into the limit:
$$\lim\limits_{x\to 0}\frac{({x-\frac{x^2}{2} + \frac{x^3}{3} + R_3(x) -x +\frac{x^2}{2})^4}}{(1-\frac{x^2}{2} + \frac{x^4}{4!} + S_3(x) -1+\frac{x^2}{2})^3}$$
Simplyfing:  
$$\lim\limits_{x\to 0}\frac{({\frac{x^3}{3} + R_3(x))^4}}{(\frac{x^4}{4!} + S_3(x))^3}$$
$R_3(x)$ and $T_3(x)$ are the remainders (with order of $3$).
We've learned in class that if you divide the remainder by the same order the limit is still approaces $0$ and I think that's the case here.
I'd be glad if you could show me how to end this exercise and explaining more about the remainder (Which approaches $0$ even if divided by a polynomial with the same order).
Thanks!
 [/INST] You might rearrange and factor some parts of the fraction appropriately to use $$\frac{\ln(1+x)}{x}\to 1,~~~\frac{\cos x-1}{x}\to 0$$ while $x\to 0$. Note that when $x\to 0$ then, we can simplify something like $\frac{x^4}{x^2}$ and get $x^2$.
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