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We're asked to find the following limit by using Taylor expansions $$lim_{xto{}0}frac{e^{3x}-sin(x)-cos(x)+ln(1-2x)}{-1+cos(5x)}$$

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 $xto{}0$, any help would be appreciated.

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 $cx^2+o(x^2)$.

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 $pminfty$ depending on the sign of $n-m$.

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?

We need only quote the numerator and denominator up to $x^2$ terms: $$lim_{xto 0}frac{1+3x+color{blue}{9x^2/2}-x-1+color{blue}{x^2/2}-2xcolor{blue}{-2x^2}+O(x^3)}{-1+1color{blue}{-25x^2/2}+O(x^3)}.$$You'll find only $x^2$ terms survive in each.