How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$

Question

How to find $$\lim_\limits{x \to 0}\frac{\cos (ax)-\cos (bx) \cos(cx)}{\sin(bx) \sin(cx)}$$

I tried using L Hospital's rule but its not working!Help please!

Answer 1

i will expand on the hint given by dr. mv. we will compute the numerator and denominators separately. we have $$\begin{align}\cos ax - \cos bx \cos cx &= 1 - a^2x^2/2 + \cdots - (1 - b^2x^2/2+\cdots)(1-c^2x^2/2+\cdots)\\&= \frac12 (b^2 + c^2 - a^2) x^2 + \cdots\\ \sin bx \sin cx &= (bx + \cdots)(cx + \cdots) = bcx^2 + \cdots\end{align}$$

therefore, $$\lim_{x \to 0}\frac{\cos ax - \cos bx \cos cx}{\sin bx \sin cx} = \frac{b^2 + c^2 - a^2}{2bc}.$$

Answer 2

HINT:

$$\cos ax =1-\frac12 (ax)^2+O(x^4)$$

and

$$\sin ax= ax+O(x^3)$$

Answer 3

$$\cos(ax)-\cos(bx)\cos(cx)=\dfrac{2\cos(ax)-2\cos(bx)\cos(cx)}2$$

$$=\dfrac{2\cos(ax)-[\cos(b-c)x+\cos(b+c)x}2$$

$$=\dfrac{1-\cos(b-c)x+1-\cos(b+c)x-2[1-\cos(ax)]}2$$

Now $\dfrac{1-\cos px}{\sin qx\sin rx}=\dfrac{1-\cos^2px}{(1+\cos px)\sin qx\sin rx}$ $=p^2\left(\dfrac{\sin px}{px}\right)^2\cdot\dfrac1{q\cdot\dfrac{\sin qx}{qx}}\cdot\dfrac1{r\cdot\dfrac{\sin rx}{rx}}\cdot\dfrac1{(1+\cos px)}$

Answer 4

Using the fact that

$ \lim_{x \to 0} \frac{\sin (\alpha x)}{x} = \alpha $

you may write

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

Then you may apply L'Hospital's rule here.

Answer 5

Before using L'Hospital, turn the products to sums

$$\frac{\cos(ax)-\cos(bx)\cos(cx)}{\sin(bx)\sin(cx)}=\frac{2\cos(ax)-\cos((b-c)x)-\cos((b+c)x)}{\cos((b-c)x)-\cos((b+c)x)}.$$

Then by repeated application $$\frac{2a\sin(ax)-(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)}{(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)},$$ and $$\frac{2a^2\cos(ax)-(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}{(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}.$$

The limit is

$$\frac{2a^2-(b-c)^2-(b+c)^2}{(b-c)^2-(b+c)^2}=-\frac{a^2-b^2-c^2}{2bc}.$$

Answer 6

One can use only the following standard limits (without L'Hospital's rule nor Taylor series) \begin{equation*} \lim_{u\rightarrow 0}\frac{1-\cos u}{u^{2}}=\frac{1}{2}\ \ \ \ \ \ \ \ ,\ \lim_{u\rightarrow 0}\cos u=\cos 0=1\ \ \ \ \ \ \ \ \lim_{u\rightarrow 0}% \frac{\sin u}{u}=1 \end{equation*} To this end, transform the original expression as follows \begin{equation*} \frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx}=\frac{\cos ax-\cos bx\cos cx}{% x^{2}}\frac{bx}{\sin bx}\frac{cx}{\sin cx}\frac{1}{bc} \end{equation*} Note that \begin{equation*} \cos bx\cos cx=(1-\cos bx)(1-\cos cx)+\cos cx+\cos bx-1 \end{equation*} then \begin{eqnarray*} \frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx} &=&\left( \frac{\cos ax-(1-\cos bx)(1-\cos cx)-\cos cx-\cos bx+1}{x^{2}}\right) \times \frac{bx}{\sin bx}\frac{cx}{\sin cx}\frac{1}{bc} \\ &=&\left( \frac{\cos ax-1}{(ax)^{2}}a^{2}+\frac{1-\cos bx}{(bx)^{2}}b^{2}+% \frac{1-\cos cx}{(cx)^{2}}c^{2}+\frac{1-\cos cx}{(cx)^{2}}(\cos bx-1)c^{2}\right) \times \frac{bx}{\sin bx} \frac{cx}{\sin cx} \frac{1}{bc}. \end{eqnarray*}

Passing to the limit as $x$ tends to $0$ and using standard limits cited above one gets \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx} &=&\left( \frac{-1}{2}a^{2}+\frac{1}{2}b^{2}+\frac{1}{2}c^{2}+\frac{1}{2}% (1-1)c^{2}\right) \times 1\cdot 1\cdot \frac{1}{bc} \\ &=&\frac{-a^{2}+b^{2}+c^{2}}{bc}. \end{eqnarray*}