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Calculate: $$\lim_{x\to0} \frac{\sin ax\cos bx}{\sin cx}$$

My attempt: $$=\lim_{x\to 0} \dfrac {\sin (ax). \cos (bx)}{\sin (cx)}$$ $$=\lim_{x\to 0} \dfrac {\sin (ax)}{ax} \times \dfrac {ax}{\sin (cx)} \times cx \times \dfrac {\cos (bx)}{cx}$$ $$=1\times ax \times 1 \times \dfrac {\cos (bx)}{cx}$$

How do I do further?

Michael Rozenberg
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pi-π
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    What is $$\cos0$$ – lab bhattacharjee Jul 18 '17 at 04:35
  • Also, do you know L'Hôpital's rule? Once you use the fact that $\lim_{u\to 0} \cos(u) = \cos(0)$, you can apply L'Hôpital's rule to the remaining limit. – Michael L. Jul 18 '17 at 04:37
  • Some nitpick. You need to assume $c\neq 0$. And further if $a=0$ then the expression vanishes so the limit is $0$. If $a\neq 0$ then your approach is OK and you just need to cancel the $x$ from $ax$ and $cx$ and note that $\cos bx\to 1$. The answer is $a/c$ and you need to keep on using $\lim_{x\to 0}$ till there is some $x$ in the expression. – Paramanand Singh Jul 18 '17 at 04:59

4 Answers4

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A bit of better than your: $$\lim_{x\rightarrow0}\frac {\sin ax\cos bx}{\sin cx}=\lim_{x\rightarrow0}\left(\cos{bx}\cdot\frac{\sin{ax}}{ax}\cdot\frac{cx}{\sin{cx}}\cdot\frac{a}{c}\right)=\frac{a}{c}$$

Michael Rozenberg
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  • There I go making things more complicated than need be. Why work so hard, when I can just multiply by $1$ in a clever way. Boy is this nicer. – A. Thomas Yerger Jul 18 '17 at 04:56
  • This is great @Michael - I would just add, in case anyone is confused, that this is using the known fact that the limit of $\frac{\sin(u)}{u}$ is $1$ as $u$ approaches $0$. – rb612 Jul 18 '17 at 06:20
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There is a $0×\infty$ indeterminate form in $1×ax×1×\frac{\cos bx}{cx}$, which would make further resolution a little harder. A more direct approach: $$\lim_{x\to0}\frac {\sin ax\cos bx}{\sin cx}=\lim_{x\to0}\frac {\sin ax}{\sin cx}$$ $$=\frac{ax-\mathcal O(x^3)}{cx-\mathcal O(x^3)}=\frac ac$$

Parcly Taxel
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There's nothing wrong with the other answer, but I like to use Taylor series for these kinds of problems, since it is easy to take the limit of a rational function. Here's a few more words about how that expansion works.

Using Taylor series, we can write the following:

$$\lim_{x \to 0} \frac{\sin (ax) \cos(bx)}{\sin(cx)} = \lim_{x \to 0}\frac{(ax - E_{\sin}(x))(1-b^2x^2+E_{\cos}(x))}{cx - E_{\sin}(x)}$$

Where the $E$ are error functions which go to $0$ as $x$ goes to $0$, and in particular, go to $0$ faster than $x$ does. Technically the top and bottom $E_{\sin}$ are different, but since they are going to $0$ in a moment anyway, we won't clutter things with any more notation.

Multiplying the limit out:

$$\lim_{x\to 0}\frac{ax - ab^2x^3 -E_{\sin}(x)(1-b^2x^2 +E_{\cos}(x))}{cx - E_{\sin}(x)}$$

Now since the error terms go to $0$ faster than $x$, we can cancel an $x$ from every term, and $\frac{E}{x} \to 0$ as $x \to 0$.

$$\lim_{x \to 0} \frac{a - ab^2x^2}{c}$$

Now the limit is obviously $a/c$, as found in the other answer.

A. Thomas Yerger
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  • Why use Taylor series when the answer is easily arrived at by using $\lim\limits_{x\to 0}\dfrac{\sin x} {x} =1$? – Paramanand Singh Jul 18 '17 at 05:02
  • @ParamanandSingh Because as I commented on the answer that uses this idea, I didn't think of it. But I did think of Taylor series. So I used them. – A. Thomas Yerger Jul 18 '17 at 05:12
  • Taylor series are also a more general tool. In this particular case, things worked out simply, but there may potentially be cases where you can't pull that particular kind of trick. – Tyg13 Jul 18 '17 at 05:14
  • @Tyg13: Agree Taylor's series are a powerful tool but the use of standard limits like $\sin x/x \to 1$ is not a trick but rather a systematic method of evaluation of limits which is much simpler to understand and does not require the idea of derivatives. One fine application of this method on a reasonably complicated limit can be seen here https://math.stackexchange.com/a/553878/72031 – Paramanand Singh Jul 18 '17 at 07:15
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$$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}$$

Using linear approximation

$$L=f(\text a)+f'(x)(x-\text a)$$

Therefore, at $x=0$ $$\sin \text ax\approx \text ax$$ $$\sin \text cx\approx \text cx$$ $$\cos bx \approx 1$$

$$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}=\lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx} \iff \lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx}=\lim_{x \rightarrow 0} \frac{\text a}{\text c}$$

$$\therefore \lim_{x \rightarrow 0} \frac{\text a}{\text c}=\frac{\text a}{\text c}$$