$$\lim_{x\to0}\frac{x(1+a\cos x)-b\sin x}{\sin^3x}=1$$
This was an example problem solved by my teacher first through L'Hopital's and then by series expansion.
I also proceeded through L'Hopital but ended with the wrong answer.
I divided the numerator and the denominator by $x$ to get:
$$\lim_{x\to0}\frac{(1+a\cos x)-b}{\sin^2x}=1$$
Since the numerator tends to zero for the limit to have a finite value, $b-a=1$
Then I divided the numerator and denominator by $x^2$ to get:
$$\lim_{x\to0}\frac{(1+a\cos x)-b}{x^2}=1$$
L'hopitals:
$$\lim_{x\to0}\frac{-a\sin x}{2x}=1$$
I get $a=-2$ but it's supposed to be $\dfrac{-5}{2}$
Which of my steps was wrong or illegal?
Best Answer
This one is wrong.
Once you split the limits using their algebra, clubbing them again may give wrong answer.
The correct method using L.H. rule will be
First multiply and divide by $x^3$ and separate $\lim_{x\to 0} \frac{x^3}{\sin^3x}=1$ to get
$$\lim_{x\to0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1\implies\lim_{x\to0}\frac{1-(b-a)\cos x -ax\sin x}{3x^2}=1$$
$$\lim_{x\to 0}\frac{1-(b-a)\cos x}{x^2}-\lim_{x\to 0}\frac{a\sin x}{x}=3$$
For the first limit to exist , $b-a=1$ which gives
$$\frac{1}{2}-a=3\implies a=-\frac{5}{2}\implies b=-\frac{3}{2} $$