# For what values of a and b does the following limit equal 0?

I understand I need to make the sum of the individual limits equal 0 - but I'm a little lost. I computed the limit of the first term to be -4/3 via L'Hospitals Rule but Wolframalpha contradicts me (http://www.wolframalpha.com/input/?i=limit+x-%3E+0+%28sin%282x%29%2F%28x%5E3%29%29).

'a' obviously should be left for last - and $b/($x^2) is some constant - so I get 0? Assuming 'b' is positive, or anything other than 0, I get that term is 0.

Thus, -4/3 + 0 + a = 0

Simple algebraic manipulating would lead me to believe a = 4/3 and then I'd plug that back in to find 'b'.

Questions

Why does Wolfram state what it does? Is this solution correct? (I don't have the answer for the problem.)

• The limit of the first summand is infinity, surely not $-4/3$. – egreg Oct 20 '13 at 23:00
• @egreg I got -4/3 from L'Hospitals rule but perhaps I screwed up.. having an infinity in there really puts a wrench in things, though. – Vaughan Hilts Oct 20 '13 at 23:02

Hint: you're correct that $a$ can be taken care of when you have found $$\lim_{x\to0}\frac{\sin(2x)+bx}{x^3}$$ that you can compute by applying l'Hôpital's theorem. But you have to compute derivatives right.

For instance, just to show your computations were wrong, $$\lim_{x\to0}\frac{\sin(2x)}{x^3}= \lim_{x\to0}\frac{2\cos(2x)}{3x^2}=\infty$$ You can't go on with l'Hôpital here, because it's not a $0/0$ form any more.

• Aw crud. No wonder, I screwed up a rule. So what do I do at that point? L'Hospitals is no good anymore and the limit is infinity... – Vaughan Hilts Oct 20 '13 at 23:07
• @VaughanHilts Consider the limit I suggested; it is a $0/0$ form, so you can apply the rule once. What do you get? Is it still $0/0$? If not, the limit is $\infty$ (because of the $3x^2$ at the denominator). If it is (but this depends on $b$), apply the rule once more. – egreg Oct 20 '13 at 23:10
• OK, so I verify THAT summand is indeed -4/3. Now, I deduce that a must be 4/3. Now, I can use that fact to find b... excellent. Thanks! – Vaughan Hilts Oct 20 '13 at 23:18

We have by the Taylor series $$\sin(2x)=2x-\frac{4}{3}x^3+o(x^3)$$ so it's clear that for $b=-2$ and $a=\frac{4}{3}$ the limit is $0$.

• Ah, I probably should have mentioned I'm interesting in doing this without a Taylor series. :) We're covering those later. – Vaughan Hilts Oct 20 '13 at 23:02
• $\ddot\smile$ Sami. +) – mrs Oct 21 '13 at 15:44

First let us put this into a better form, with one variable term and one constant term: $$\lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} + a = 0$$ well if we evaluate the limit using L'Hopitals we get: $$\lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} = \lim\limits_{x \to 0} \frac{2\cos(2x) + b}{3x^2}$$ $$\lim\limits_{x \to 0} \frac{2\cos(2x) + b}{3x^2} = \frac{2+b}{0} = (2+b) \infty$$ Since this technically equals infinity in all cases except when $(2+b) = 0$, we make the $b = -2$ so that our limit comes out to be zero. Now let's figure out the $a$

First I want to plug in the $b$ back into our original equation: $$\lim\limits_{x \to 0} \frac{\sin(2x) + bx}{x^3} + a = \lim\limits_{x \to 0} \frac{\sin(2x) - 2x}{x^3} + a = 0$$ then let's take the limit $$\lim\limits_{x \to 0} \frac{\sin(2x) - 2x}{x^3} + a = \lim\limits_{x \to 0} \frac{2\cos(2x) - 2}{3x^2} + a = \frac{2}{3}\lim\limits_{x \to 0} \frac{\cos(2x) - 1}{x^2} + a$$ and we can apply the L'Hospitals rule again: $$\frac{2}{3}\lim\limits_{x \to 0} \frac{\cos(2x) - 1}{x^2} + a = \frac{2}{3} \lim\limits_{x \to 0} \frac{-2\sin(2x)}{2x} + a = \frac{2}{3} \lim\limits_{x \to 0} \frac{-\sin(2x)}{x} + a$$ and once again with L'Hospitals: $$\frac{2}{3} \lim\limits_{x \to 0} \frac{-\sin(2x)}{x} +a = \frac{2}{3} \lim\limits_{x \to 0} {-2\cos(2x)} + a = \frac{-4}{3} + a$$ If we then set this equal to zero we get $a = \frac{4}{3}$ And so our final answer is: $$\lim\limits_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}= 0 , a = \frac{4}{3}, b = -2$$

• Typo on the last line - you say a = 3/4 but really it should be 4/3. Other than that, solid answer and what I ended up getting after some other help. Thanks a lot of this solution! – Vaughan Hilts Oct 20 '13 at 23:43