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Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$ What is the value of $a+b$

My try

$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$

$\lim\limits_{x\to 0} (\frac {(1+a×cos(x)}{x^2}-\frac {b}{x^2})=1$

$\lim\limits_{x\to 0} \frac {1+a\cos x-bx}{x^2}=1$

Apply L'Hôpital's rule: $\lim\limits_{x\to 0} \frac {-a\cos x-b}{2x}=1$

Apply L'Hôpital's rule again: $\lim\limits_{x\to 0} \frac {-a\sin x}{2}=1$ $\to$ $a=-2$

Is my approach right?

user373141
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  • I'd recommend checking when you are allowed to use L'Hospital's rule, mainly near the end. Usually it requires a $0/0$ form, but that is not necessarily the case.... – Simply Beautiful Art Aug 04 '17 at 12:00

4 Answers4

5

Series methods are always useful.

$$\cos(x)=1-\frac12x^2+\mathcal O(x^3)\\1+a\cos(x)=1+a-\frac a2x^2+\mathcal O(x^3)\\x(1+a\cos(x))=(1+a)x-\frac a2x^3+\mathcal O(x^4)\\\sin(x)=x-\frac16x^3+\mathcal O(x^4)\\x(1+a\cos(x))-b\sin(x)=(1+a-b)x+\frac{b-3a}6x^3+\mathcal O(x^4)\\\frac{x(1+a\cos(x))-b\sin(x)}{x^3}=\frac{1+a-b}{x^2}+\frac{b-3a}6+\mathcal O(x)$$

For this to go to $1$ as $x\to0$, we need $1+a-b=0$ and $\frac{b-3a}6=1$, which gives us

$$\begin{cases}b-3a=6\\1+a-b=0\end{cases}\implies a=-\frac52,b=-\frac32\\\implies a+b=-4$$

Simply Beautiful Art
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5

It is:

$$L=\lim_{x\to 0} \frac {x+ax\cos x-b\sin x}{x^3}\stackrel{LR}=\lim_{x\to 0} \frac {1+a\cos x-ax\sin x-b\cos x}{3x^2}\stackrel{1+a-b=0; LR}=\lim_{x\to 0} \frac {-a\sin x-a\sin x-ax\cos x+b\sin x}{6x}\stackrel{LR}=\lim_{x\to 0} \frac {-a\cos x-a\cos x-a\cos x+ax\sin x+b\cos x}{6}=1 \Rightarrow -3a+b=6.$$ Hence: $$\begin{cases}1+a-b=0 \\ -3a+b=6\end{cases} \Rightarrow a+b=-\frac52-\frac32=-4.$$

farruhota
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  • How do you know that limits after the application of L'Hospital's Rule exist? You are only given that the original limit exists. Surprisingly this invalid usage of L'Hospital's Rule is given upvotes without much thought. – Paramanand Singh Aug 04 '17 at 13:56
  • @ParamanandSingh, thank you for your critique, but each use of the L'Hospital's rule is backed up by the indeterminate form of $0/0$. Hopefully you noticed the condition $1+a-b=0$ before the application of the LR. The existence of the limit is a key throughout – farruhota Aug 04 '17 at 14:43
  • Perhaps you need to revisit the L'Hospital's Rule. If we restrict to the form $0/0$ then it says that if $f'/g'\to L$ then $f/g \to L$. Here you are given that $f/g\to L$ and $f\to 0,g\to 0$. From this given information you can't conclude anything about the limit of $f'/g'$. In general I have found that rarely do people understand the exact nature and working of this infamous L'Hospital's Rule. – Paramanand Singh Aug 04 '17 at 15:34
  • Thus all your implications are backwards and in essence you have proved that if $a=-5/2,b=-3/2$ then the limit in question is $1$. You are supposed to prove the converse "limit in question is $1\Rightarrow a=-5/2,b=-3/2$. – Paramanand Singh Aug 04 '17 at 15:36
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    From your last comment it appears that you have taken my feedback in positive manner (this is uncommon especially for negative feedback). Thanks for the same! The issue in your method is difficult to detect (because it gives correct answer) and you will need to check very thoroughly the hypotheses of L'Hospital's Rule to understand my statements. – Paramanand Singh Aug 04 '17 at 15:45
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    It should be possible however to supplement this answer with some further arguments (like consider what happens if $1+a-b\neq 0$) to make it fully correct. – Paramanand Singh Aug 04 '17 at 15:55
  • @ParamanandSingh, good points, I agree if $a=-5/2,b=-3/2$, then $f'''/g'''$, then $f''/g''$, then $f'/g'$, then $f/g$ exist, which is in compliance with the L'opital rule. It is indeed solved backwards. It is required to find the parameters $a$ and $b$ if or while the limit exists. – farruhota Aug 04 '17 at 16:40
  • Ok for the benefit of everyone I will fix the issue. Just after the first application of L'Hospital's Rule let's consider that $1+a-b\neq 0$. Then the limit of the resulting expression is $(\text{sign of} (1+a-b))\cdot\infty$. And then the original limit must also diverge which is not the case. Hence we must have $1+a-b=0$. And then the expression becomes $$\frac{1-\cos x} {3x^{2}}-\frac{a}{3}\frac{\sin x} {x} $$ which tends to $(1/6)-(a/3)$ and this should be equal to $1$. Thus $a=-5/2,b=a+1=-3/2$. You can also see that further applications of L'Hospital's Rule are not really needed. – Paramanand Singh Aug 05 '17 at 02:27
  • Thus $1+a-b$ becomes $0$ $\textbf{not} $ because we want the expression after L'Hospital's Rule to be in form $0/0$, but because the original limit exists finitely. And this way it works as a forward implication. Such analysis of hypotheses/conclusion of L'Hospital's Rule is typically a must before and after its application. – Paramanand Singh Aug 05 '17 at 02:41
  • @ParamanandSingh, I think you are just interpreting my shorthand $1+a-b=0$. Of course, the limit after the first use of LR has two options: exist or not exist. If it exists, it must be 1. If not, again two options: the original limit does not exist or it is indeterminate form of either $0/0$ or $\infty/\infty$. Since $1+a-b$ is constant, the only possibility is $0/0$. – farruhota Aug 05 '17 at 04:16
  • No! It either exists finitely (in which case it must equal $1$), or it diverges (to $\pm\infty$) in which case original also does the same so this is ruled out, or the limit does not exist (oscillates finitely or infinitely) and this case is also ruled out depending on values of $1+a-b$. Also note that the third case of oscillation is possible even if original limit exists. It is not necessary that it has to be of the form $0/0$ or $\infty/\infty$. Such forms are necessary before application of L'Hospital's Rule and not after. – Paramanand Singh Aug 05 '17 at 04:31
2

As $x(1+a\cos x)-b\sin x\sim (1+a-b)x$ as $x\to0$ then the given limit can only exist when $b=1+a$. A series expansion then gives $$x(1+a\cos x)-(1+a)\sin x\sim-\frac{a}2x^3+(1+a)\frac{x^3}6.$$ One needs then $(1+a)/6-a/2=1$ for the given limit to equal $1$. Now one can obtain $a$ and $b$.

Angina Seng
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We need two limits below (which are easily obtained and the second one necessitates the use of Taylor series or L'Hospital's Rule) $$\lim_{x\to 0}\frac{1-\cos x} {x^{2}}=\frac{1}{2},\,\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}$$ The limit in question can be written as $$\lim_{x\to 0}(a+1)\cdot\frac{x-\sin x} {x^{3}}-a\frac{1-\cos x} {x^{2}}-\frac{(b-a-1)\sin x} {x^{3}}=1$$ This means that $$\lim_{x\to 0}\frac{(b-a-1)\sin x}{x^{3}}=1+\frac{a}{2}-\frac{a+1}{6}\tag{1}$$ or (upon multiplication by $x^{2}$) $$\lim_{x\to 0}\frac{(b-a-1)\sin x} {x} =0$$ ie $b-a-1=0$. Using this in $(1)$ we get $$1+\frac{a}{2}-\frac{a+1}{6}=0$$ ie $a=-5/2,b=a+1=-3/2,a+b=-4$.

Your approach has a fundamental issue when you replace $(b\sin x) /x^{3}$ by $b/x^{2}$. This is not allowed by any theorem on limits and is an invalid step in the evaluation of a limit.

Paramanand Singh
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  • $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ – user373141 Aug 04 '17 at 14:59
  • @prayersmith: that equation means that wherever you see $\lim\limits_{x\to 0} \dfrac {\sin x} {x} $ you can put $1$ in place of it. It does not mean that wherever you see $\dfrac{\sin x} {x} $ you replace it by $1$. – Paramanand Singh Aug 04 '17 at 15:30
  • @prayersmith: the flaw in your approach is common and I have discussed elsewhere about it on this website in more detail. Please have a look at this answer: https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Aug 04 '17 at 15:49
  • but i can distribute the limit over the difference; what's wrong with that? – user373141 Aug 04 '17 at 16:04
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    @prayersmith : you can do that only when you know that limits exist. The limit of $(b\sin x) /x^{3}$ does not exist. My linked answer in previous comments mentions the condition under limits can be distributed with respect to $+, -$ and the condition is that one of the limits must exist finitely. – Paramanand Singh Aug 04 '17 at 16:06