3

Question : -

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

My attempt :-

$$\lim_{x \to 0} \frac{2* \sin ax * \cos bx + 2*\sin ax * \cos (ax)}{(\cos bx - \cos ax)*(\cos bx +\cos ax)}$$ $$\lim_{x \to 0} \frac{2* \sin ax * (\cos bx + \cos (ax))}{(\cos bx - \cos ax)*(\cos bx +\cos ax)}$$ $$\lim_{x \to 0} \frac{2* \sin ax}{(\cos bx - \cos ax)}$$ $$\lim_{x \to 0} \frac{2* \sin ax}{(-2 * \sin {bx-ax\over2} * \sin {bx+ax\over2})}$$ $$-\lim_{x \to 0} \frac{{\sin ax \over ax} * {ax }}{\frac{\sin {bx-ax\over2} * \sin {bx+ax\over2}}{ {bx+ax\over2} * {bx-ax\over2}} * {bx+ax\over2} * {bx-ax\over2}}$$ $$-\lim_{x \to 0} \frac{ax}{{bx+ax\over2} * {bx-ax\over2}}$$ $$-\lim_{x \to 0} \frac{ax}{{(bx)^2 - (ax)^2\over 4}}$$ $$\lim_{x \to 0} \frac{4 * ax}{(ax)^2 -(bx)^2}$$ $$\lim_{x \to 0} \frac{4 * a}{x (a^2 - b^2)}$$

There i hit the dead end, i can't do anything now.

Can anyone please tell me what i have done incorrect here ?

Sorry but i can't confirm the legitimacy of the question, i just found it on net and there is no answer given.

  • Well assuming that your attempt is correct(I did not check it)...you will end up getting the answer as, "Limit does not exist", simply because you last statement reads as $\lim_{x\to 0}\frac{k}{x}$(where $k$ is a real constant). This will go to $\infty$. – Coherent Sheaf Jun 11 '16 at 15:11
  • @Dr.MV So there is no way of doing this question ? –  Jun 11 '16 at 15:14
  • 2
    The limit $\lim_{x\to 0}\frac1x$ does not exist since the limits from the right and left are opposite in sign. So, the only case for which the limit of interest exists in for $a=0$. And that does answer the question. Your analysis was solid! – Mark Viola Jun 11 '16 at 15:15

2 Answers2

2

I'll assume that $a^2-b^2\ne0$ and $a\ne0$.

Notice that $\cos^2(bx)-\cos^2(ax)=(\cos(bx)+\cos(ax))(\cos(bx)-\cos(ax))$ and that the first factor has limit $2$, so it can be set away momentarily. Then $$ \cos(bx)-\cos(ax)=1-\frac{b^2x^2}{2}+o(x^2)-1+\frac{a^2x^2}{2}+o(x^2) =\frac{a^2-b^2}{2}x^2+o(x^2) $$ and so the denominator becomes $$ (a^2-b^2)x^2+o(x^2) $$ because the other factor is $2+o(1)$.

The numerator is $$ (a+b)x+(a-b)x+2ax+o(x)=4ax+o(x) $$ and so the limit is infinite (different from the right than from the left).

Your computation seems correct, as you arrive essentially at the same point.

If $a=0$, the numerator becomes identically $0$, so the limit is $0$.

If $a^2=b^2$, then the denominator vanishes identically, so the limit is meaningless.

egreg
  • 238,574
  • What do you mean by meaningless limit ? really sorry. –  Jun 11 '16 at 15:19
  • 1
    @ritwiksinha If the denominator is everywhere $0$, the function is nowhere defined, so you can't consider a limit for it. – egreg Jun 11 '16 at 15:24
1

HINT:

$$\cos^2bx-\cos^2ax=\sin^2ax-\sin^2bx=\sin(a+b)x\sin(a-b)x$$ using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

and we know $\lim_{h\to0}\dfrac{\sin h}h=1$

Clearly the limit reduces to $$\dfrac{2a}{(a+b)(a-b)}\lim_{x\to0}\dfrac1x$$ which is non-existent

Community
  • 1
lab bhattacharjee
  • 274,582