2

if :

$$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$

then $a+b=?$

Without the use of the L'Hôspital's Rule

My Try :

$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$

$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)+\sin 3x}{x^3} \right)=0$$

$$\lim_{ x \to 0 }x(a+bx^2)+\sin 3x=0 $$

now ?

Almot1960
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  • See the answers of http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion for $$\dfrac{x-\sin x}{x^3}$$ – lab bhattacharjee Feb 25 '17 at 09:21

4 Answers4

2

With Taylor expansion $$ \begin{aligned} \lim _{x\to 0}\left(\frac{\sin \left(3x\right)+ax+bx^3}{x^3}\right) & = \lim _{x\to \:0}\left(\frac{\left(3x-\frac{9}{2}x^3+\frac{81}{40}x^5+o\left(x^5\right)\right)+ax+bx^3}{x^3}\right) \\ & = \lim _{x\to \:0}\left(\frac{81x^2}{40}\right) = 0 \end{aligned} $$ Then $a$ and $b$ must have the following values $$a = -3, b = 9/2$$ So: $$\lim _{x\to \:0}\left(\frac{\sin \left(3x\right)-3x+\frac{9}{2}x^3}{x^3}\right) = 0$$

Amarildo
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1

The MacLaurin expansion of $\sin3x$ up to order $3$ is $$\sin3x = 3x - \frac{9x^3}{2} + \mathcal O(x^5)$$ Plugging in the limit we obtain $$\lim_{x \to 0} \frac{(3 + a)x + (b - \frac92)x^3}{x^3} =0$$

If the numerator is not zero then the limit is infinite, because the denominator goes to $0$ as order $3$. Therefore it has to be $$\left\{\begin{align*} a &= -3\\ b &= \frac92 \end{align*}\right.$$

rubik
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1

Why not to use Taylor series $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin(3x)=3 x-\frac{9 x^3}{2}+\frac{81 x^5}{40}+O\left(x^6\right)$$ $$\frac{\sin 3x}{x^3}+\frac{a}{x^2}+b =\frac{a+3}{x^2}+\left(b-\frac{9}{2}\right)+\frac{81 x^2}{40}+O\left(x^3\right)$$ Then $???$

Claude Leibovici
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1

We need to use the limit $$\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}\tag{1}$$ which can be easily established either via L'Hospital's Rule or Taylor's theorem. Now using $(1)$ we have $$\frac{\sin 3x - 3x}{x^{3}}\to - \frac{9}{2}\tag{2}$$ and hence $$\frac{\sin 3x}{x^{3}}+\frac{a}{x^{2}}+b\to 0\tag{3}$$ is equivalent to (via subtracting equation $(2)$ from equation $(3)$) $$\frac{3+a}{x^{2}}+b\to \frac{9}{2}\tag{4}$$ or $$\frac{3+a+bx^{2}}{x^{2}}\to \frac{9}{2}\tag{5}$$ It is now clear that $3+a+bx^{2}\to 0$ so that $a=-3$. And then $b=9/2$ follows from the penultimate equation. Note that everything apart from limit $(1)$ is based on algebra of limits.

Paramanand Singh
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