Evaluating the limit using Taylor Series

Question

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.stack.imgur.com/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.

Answer 1

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?

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

Answer 2

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.

Answer 3

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

$$-\frac{6}{25}.$$

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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 $n-m$.

Answer 4

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?