Evaluating the limit using Taylor Series

analysiscalculuslimitstaylor expansion

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://i.sstatic.net/I9Qd0.jpg (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.

Best Answer

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

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