Evaluating the limit using Taylor Series We're asked to find the following limit by using Taylor expansions $$\lim_{x\to{}0}\frac{e^{3x}-\sin(x)-\cos(x)+\ln(1-2x)}{-1+\cos(5x)}$$
My Attempt:
Expressing $e^{3x}$, $\sin(x)$, $\cos(x)$, $\ln(1-2x)$ and $\cos(5x)$ in their respective taylor expansions yielded the following monstrous fraction, https://imgur.com/a/xGyfIyL (Picture size too big to be uploaded here for some reason, plus fraction too large to be expressed in the space given :/) But anyways, I can't seem to factorize this thing and evaluate the limit as $x\to{}0$, any help would be appreciated.
 A: HINT
By Taylor's expansion, term by term, we have that


*

*$e^{3x}=1+3x+\frac92x^2+o(x^2)$

*$\sin x =x+o(x^2)$

*$\cos x = 1-\frac12 x^2+o(x^2)$

*$\log(1-2x)=-2x-2x^2+o(x^2)$

*$\cos (5x) = 1-\frac{25}2 x^2+o(x^2)$
and then
$$\frac{e^{3x}-\sin(x)-\cos(x)+\ln(1-2x)}{-1+\cos(5x)}=\frac{1+3x+\frac92x^2-x-1+\frac12x^2-2x-2x^2+o(x^2)}{-1+1-\frac{25}2x^2+o(x^2)}$$
Can you conclude from here?

Edit for a remark
The main point with Taylor's expansion is to guess the correct order to use for the expansion and there is not general a rule to be sure about the order to use.
In the doubt, we could decide to start with the first order to obtain $$\frac{e^{3x}-\sin(x)-\cos(x)+\ln(1-2x)}{-1+\cos(5x)}=\frac{1+3x-x-1-2x+o(x)}{-1+1+o(x)}$$ which is an indeterminate form $\frac 0 0$. 
When this happen it means that the order used is not sufficient. Then we can try with the order $2$ which indeed works fine. Of course also all the orders greater than $2$ work fine. But the second order suffices, how we can easily guess, in that case, looking at the denominator which is in the form $cx^2+o(x^2)$.
A: We need only quote the numerator and denominator up to $x^2$ terms: $$\lim_{x\to 0}\frac{1+3x+\color{blue}{9x^2/2}-x-1+\color{blue}{x^2/2}-2x\color{blue}{-2x^2}+O(x^3)}{-1+1\color{blue}{-25x^2/2}+O(x^3)}.$$You'll find only $x^2$ terms survive in each.
A: There is nothing monstrous in that computation. If you want a compact resolution, compute separately the coefficients in increasing order (numerator/denominator).


*

*constant terms: $1-1=0\ /\ -1+1=0$;

*linear terms: $3-1-2=0\ /\ 0$;

*quadratic terms: $\dfrac92+\dfrac12-2=\dfrac{6}{2}\ / -\dfrac{25}2$.
As the first nonzero coefficients are of the same order, the limit is finite and is the ratio
$$-\frac{6}{25}.$$

The trick is to obtain a fraction like
$$\frac{ax^n+\text{higher order terms}}{bx^m+\text{higher order terms}}=x^{n-m}\frac{a+\text{higher order terms}}{b+\text{higher order terms}}$$ which tends to $0,\dfrac ab$ or $\pm\infty$ depending on the sign of $n-m$.
A: From your picture, it seems an expansion at order $2$ will do. All functions should be expanded at the same order, using, say Taylor-Young's formula. Thus


*

*$\mathrm e^{3x}=1+3x+\frac92x^2+o(x^2)$,

*$\sin x =x+o(x^2)$,

*$\cos x=1-\frac12 x^2+o(x^2)$,

*$\ln(1-2x)=-2x-\frac42 x^2+o(x^2)$
Thus the numerator is
$$N(x)=1+3x+\frac92x^2-x-1+\frac12 x^2-2x-2 x^2+o(x^2)=3 x^2+o(x^2).$$
Can you proceed?

