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Let $\alpha, \beta \in \mathbb{R}$ be such that $$\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$$ then what is the value of $6(\alpha+\beta)$

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7 Answers 7

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Applying l'Hospital:

$$\lim_{x\to0}\frac{x^2\sin\beta x}{\alpha x-\sin x}=\lim_{x\to0}\frac{2x\sin\beta x+\beta x^2\cos\beta x}{\alpha-\cos x}$$

Since the limit in the numerator is zero, it must be the same in the denominator, so $\;\alpha=1\;$ ...Can you take it from here?

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  • $\begingroup$ @ResearchEngineer No intuition at all: we are given that the limit equals $\;1\;$ . If $\;a\neq1\;$ , then the limit is zero over non-zero = zero...! Thus it must be that also the denominator vanishes and thus we can use l'Hospital once again. $\endgroup$
    – DonAntonio
    Apr 4, 2017 at 20:08
  • $\begingroup$ @ResearchEngineer Because $\;\alpha-\cos x\xrightarrow[x\to0]{}\alpha-1\;$ , and since this is zero then $\;\alpha=1\;$ . $\endgroup$
    – DonAntonio
    Apr 4, 2017 at 20:11
  • $\begingroup$ @ResearchEngineer In the right hand of my solution, the numerator clearly tends to zero when $\;x\to0\;$ , right? If $\;a\neq1\;$, then the whole overall limit is $\;\frac0{\alpha-1}=0\;$ , and this contradicts what you're given! Thus, it must be that $\;\alpha=1\;$ so that the denominator on the right side equals zero, and thus you can do l'Hospitale once again (in fact twice) . $\endgroup$
    – DonAntonio
    Apr 4, 2017 at 20:17
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You can rewrite your function as $$ \frac{\beta x^3}{\alpha x-\sin x}\frac{\sin(\beta x)}{\beta x} $$ If $\alpha\ne 1$, this also becomes $$ \frac{\beta x^2}{\alpha-\dfrac{\sin x}{x}}\frac{\sin(\beta x)}{\beta x} $$ Note that the second factor has limit $1$ and the first factor has limit $0$ (the numerator has limit $0$; the denominator has limit $\alpha-1\ne0$).

Thus if the limit has to be $1$, we need $\alpha=1$.

Now it should be well known that $$ \lim_{x\to0}\frac{x-\sin x}{x^3}=\frac{1}{6} $$ (apply l'Hôpital or a Taylor expansion), so we have $$ \lim_{x\to0} \frac{\beta}{\dfrac{x-\sin x}{x^3}} \frac{\sin(\beta x)}{\beta x} =6\beta $$ Now it's just an easy computation.

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$$ \begin{aligned} \lim _{x\to 0}\left(\frac{x^2\sin \left(\beta \:x\right)}{\alpha \:x-\sin \:x}\right) & = \lim _{x\to 0}\left(\frac{x^2\left(\beta x+o\left(x\right)\right)}{\alpha \:x-\left(x-\frac{x^3}{3!}+o(x^3)\right)}\right) \\& = \beta \:\cdot \lim _{x\to 0}\left(\frac{6x^2}{6α+x^2-6}\right) (\text{so } \color{red}{\beta \ne 0}) \\& = \beta \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = 6\beta \text{ (with } \color{red}{\alpha = 1}) \\& = \color{red}{\frac{1}{6}} \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = \color{blue}{1} \text{ (with } \color{red}{\beta = \frac{1}{6}}) \end{aligned} $$

So, with $$\alpha = 1, \beta = \frac{1}{6}$$ $$6(\alpha + \beta)=6\left(1+\frac{1}{6}\right) = \color{gold}{7}$$

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HINT:

As $x\to0$, $\sin(\beta x)=\beta x+O(x^3)$ and $\sin(x)=x-\frac16 x^3+O(x^5)$

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  • $\begingroup$ Of course this did not help: you're only interested in the final result. $\endgroup$
    – DonAntonio
    Apr 4, 2017 at 20:21
  • $\begingroup$ @researchengineer What did you try to do with the hints? $\endgroup$
    – Mark Viola
    Apr 4, 2017 at 20:41
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Apply L'Hospital's rule repeatedly. Doing so just once shows that $\alpha$ must be $1$. Then doing so two more times shows that $6 \beta$ is the limit, i.e., $6 \beta = 1$, or $\beta = \tfrac{1}{6}$. Thus, $6(\alpha+\beta) = 6(1 + \tfrac{1}{6}) = 7$.

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    $\begingroup$ @ResearchEngineer It is the same as mine, but he is giving you the final result without any further explanation. Apparently you didn't want to learn stuff, only to get the result. You could have said that in your question... $\endgroup$
    – DonAntonio
    Apr 4, 2017 at 20:13
  • $\begingroup$ I don't mind if the solution is posted. Please check it, verifying the final answer on your own. $\endgroup$
    – Mark Twain
    Apr 4, 2017 at 20:17
  • $\begingroup$ @ResearchEngineer Have it as you want: you don't even understand a trivial step in my proof, much less you can understand how come $\;\beta=1/6\;$ . You were only interested in the final result, not in how to achieve it. Anyone could have told you from the beginning that $\;l\alpha=1,\,\beta=1/6\;$ ...but if you write that in an exam or work, any decent college/university will grade you zero points. In this answer you were given the final result only after being hinted that l'Hospital must be used...in fact, thrice. What "clear explanation" are you talking about?? $\endgroup$
    – DonAntonio
    Apr 4, 2017 at 20:20
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    $\begingroup$ @ResearchEngineer Of course "good" I am not your professor: in any of my classes you'd have to show your work and explain it, and not only show the final result. You can congratulate yourself again I am not your professor: you'd flunk any of my courses . $\endgroup$
    – DonAntonio
    Apr 4, 2017 at 20:24
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I tried to not use L'Hospitale to make the result more explicative:

First you should note that $\beta$ is not zero, because the resulting limit is not one.

Second:

$$\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x}=1 \iff \lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x}-1=0$$ $$\iff \lim_{x\to 0} \frac{-x α + \sin(x) + x^2 \sin(x β)}{x α - \sin(x)}=0$$ $$\iff \lim_{x\to 0} \frac{-α + \frac{\sin(x)}{x} + x \sin(x β)}{α - \frac{\sin(x)}{x}}=0$$

$$\iff \lim_{x\to 0} \frac{-α + \frac{\sin(x)}{x} + x^2β \frac{\sin(x β)}{\beta x}}{α - \frac{\sin(x)}{x}}=0$$

if $\alpha$ is something different from 1 then the limit would not be $0$.

I think that the rest you should complete it.

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Since the given limit $$\lim_{x \to 0}\frac{x^{2}\sin \beta x}{\alpha x - \sin x} = 1$$ is non-zero, it follows that both the numerator and denominator of the expression under limit operation are non-zero as $x \to 0$. Hence $\beta \neq 0$. Now taking reciprocals we see that $$\lim_{x \to 0}\frac{\alpha x - \sin x}{x^{2}\sin \beta x} = 1$$ or $$\lim_{x \to 0}\frac{\alpha x - \sin x}{\beta x^{3}}\cdot\frac{\beta x}{\sin \beta x} = 1$$ or $$\lim_{x \to 0}\frac{\alpha x - \sin x}{x^{3}} = \beta\tag{1}$$ We next use the well known limit $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \frac{1}{6}\tag{2}$$ (easily proved by Taylor's series expansion or L'Hospital's Rule). Subtracting $(2)$ from $(1)$ we get $$\lim_{x \to 0}\frac{\alpha - 1}{x^{2}} = \beta - \frac{1}{6}\tag{3}$$ and then $$\alpha - 1 = \lim_{x \to 0}\frac{\alpha - 1}{x^{2}}\cdot x^{2} = \left(\beta - \frac{1}{6}\right)\cdot 0 = 0$$ so we can see that $\alpha = 1$ and from $(3)$ we now get $\beta = 1/6$. Thus $6(\alpha + \beta) = 7$.


Most answers here try to deduce $\alpha = 1$ by using contradiction. It is much easier to use algebra of limits directly to reach the conclusion $\alpha = 1$.

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  • $\begingroup$ @ResearchEngineer: I don't know how the website math9to12.org works. You can try to copy paste this answer there. I don't have any problem with this. $\endgroup$
    – Paramanand Singh
    Apr 6, 2017 at 10:24

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