Find the value of $a+b$ if $\lim _{x\to 0}\left(\frac{\tan\left(2x\right)\:-2\sin\left(ax\right)}{x\left(1-\cos\left(2x\right)\right)}\right)=b$

Question

Find the value of $a+b$ if $$\lim _{x\to 0}\left(\frac{\tan\left(2x\right)\:-2\sin\left(ax\right)}{x\left(1-\cos\left(2x\right)\right)}\right)=b$$

I tried plugging in $x=0$ in the function which gives a $\frac{0}{0}$ indeterminate form,So used L'Hopital's Rule.The simplified final answer being

$$\lim _{x\to 0}\frac{\sec^2\left(x\right)-a\cos\left(ax\right)}{\sin^2\left(x\right)+x\sin\left(2x\right)}$$

On again plugging in $x=0$, I get $\frac{1-a}{0}$ (not in indeterminate form)and therefore cannot use the L'hopital's rule.I don't think differentiating again might be a good idea (i.e if its possible) as the degree of $a$ becomes 2 and it is $a+b$ that we wish to find.

So how do I go further with this? Thanks in advance.

Answer 1

Use the small-$y$ results$$\tan y=y+\frac13y^3+o(y^3),\,\sin y=y-\frac16y^3+o(y^3).$$As $x\to0$,$$\frac{\tan(2x)-2\sin ax}{x(1-\cos 2x)}\sim\frac{2(1-a)x+(8+a^3)x^3/3+o(x^3)}{2x^3+o(x^3)}.$$If this has finite limit, $a=1$ so the limit is $b=\frac{8+a^3}{6}=\frac32$, whence $a+b=\frac52$.

Answer 2

It is possible to do this without L'Hopital or Taylor series. It is helpful to know the limits

$$\lim_{x\to 0} \frac{\tan x}{x} = 1 \hspace{40 pt} \lim_{x\to 0} \frac{\sin x}{x} = 1$$

This means that the only $a$ that makes the numerator vanish quadratically (in resonance) is $a=1$. Rearranging the limit expression with trig identities we have that

$$\frac{\tan 2x - 2\sin x}{x(1-\cos 2x)} = \frac{\frac{2\sin x\cos x}{\cos 2x}-2\sin x}{2x\sin^2 x} = \frac{2}{\cos 2x} \frac{\cos^2\frac{x}{2} - \cos^2 x}{x\sin x}$$

$$= \frac{2\left(\cos\frac{x}{2} + \cos x \right)}{\cos 2x}\cdot\frac{x}{\sin x} \cdot \frac{\cos\frac{x}{2} - \cos x}{x^2} $$

The limit of the left terms are all finite numbers, leaving us only to evaluate the right-most term. With angle addition formulas we get that

$$\frac{\cos\frac{x}{2} - \cos x}{x^2} = \frac{2 \sin \frac{3x}{4} \sin \frac{x}{4}}{x^2} = \frac{3}{8} \cdot \frac{\sin \frac{3x}{4}}{\frac{3x}{4}} \cdot\frac{\sin \frac{x}{4}}{\frac{x}{4}}$$

which means that the original limit is

$$\lim_{x\to 0} \frac{3}{4} \frac{\left(\cos\frac{x}{2} + \cos x \right)}{\cos 2x}\cdot\frac{x}{\sin x} \cdot \frac{\sin \frac{3x}{4}}{\frac{3x}{4}} \cdot\frac{\sin \frac{x}{4}}{\frac{x}{4}} = \frac{3}{2}$$

Therefore $a+b = \frac{5}{2}$