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.

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 $n-m$.
