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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.

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    $\begingroup$ A hint is that only the leading terms are going to matter. Can you see what happens to the constant, $x$, and $x^2$ terms in the numerator and denominator? $\endgroup$ – user113102 Dec 7 '18 at 21:23
  • $\begingroup$ As @gimusi notes, you do not need the explicit Taylor series for the fraction. You can answer the question with some algebra on the leading terms for numerator and denominator. $\endgroup$ – Ethan Bolker Dec 7 '18 at 21:24
  • $\begingroup$ The main point is to guess the correct order for the expansion. 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 looking at the denominator. $\endgroup$ – user Dec 7 '18 at 21:46
  • $\begingroup$ I see now how I can go about evaluating the limit itself although I still find the concept a little bit vague, as in considering a specific order for the expansion and then applying it for all the functions while leaving the rest of the series behind, it'll probably start making more sense in a few more examples. Thanks for the help :) $\endgroup$ – kareem bokai Dec 7 '18 at 21:52
  • $\begingroup$ @kareembokai Yes of course you need only to practice to get confident with the method. The main rule is that all the function must be expanded to the same order. Note also that we can get an uncorrect evaluation when the order is too low than the minimum requested but not when the order is higher (but it requires more effort and work). Then you need to practice a lot with that in order to becomemore and more confident. $\endgroup$ – user Dec 7 '18 at 21:59
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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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  • $\begingroup$ I do see that you've ended up with the leading terms, however I still don't see how you managed to get rid of all the other terms so quickly $\endgroup$ – kareem bokai Dec 7 '18 at 21:31
  • $\begingroup$ @kareembokai I add something on that! $\endgroup$ – user Dec 7 '18 at 21:32
  • $\begingroup$ @kareembokai Let me know whether now it is clear or not. In case do not hesitate to ask for any clarification. $\endgroup$ – user Dec 7 '18 at 21:40
  • $\begingroup$ In general it's better to start with the term that appears simpler, in this case the denominator, which is easily seen to have order $2$. $\endgroup$ – egreg Dec 7 '18 at 22:13
  • $\begingroup$ @egreg Yes I agree in that case is quite simple, is some other cases for a student new in the topic it can be not so straightforward to see, but after a while and some tons of limits it becomes a simple task for usual limits. $\endgroup$ – user Dec 7 '18 at 22:16
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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 $n-m$.

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    $\begingroup$ Nice way to simplify even if it requires a few of experience to be handle in that way. $\endgroup$ – user Dec 7 '18 at 21:43
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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?
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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.

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