Find value of $a$ and $b$

Question

$f(x)$ is the function such that $$\lim_{x\to 0} \frac {f(x)}{x}=1$$. If $$\lim_{x\to 0} \frac {x(1+a\cos x) -b\sin x}{(f(x))^3}=1$$ then find a, b.

Using L'Hospital I have found that $f'(0)=1$ but I am not able to use that information further. Moreover I also think that $f(0)$ must be $0$ for application of L'Hospital.

Any new methods and hints would be appreciated. If anyone comes up with a solution without use of L'Hospital then it would be the best.

Answer 1

Hint

$$ \frac {x(1+a\cos x) -b\sin x}{(f(x))^3}= \frac {x(1+a\cos x) -b\sin x}{x^3}\frac{x^3}{(f(x))^3}$$ and $$\lim_{x\to 0} \frac{x^3}{(f(x))^3}=1$$

Answer 2

• Apply LH rule to given expression
• See that of the resultant expression, for limit to exist, Numerator-> 0 (since Denominator->0)

Use the above 2 facts repeatedly to obtain equations in a and b.

A solution using series expansion is also possible.

Answer 3

Using the expansion of $\sin$ and $\cos$ you have: $$x(1+a\cos(x))-b\sin(x) = x \left(1+a \left(1-\frac{x^2}{2}+O(x^4) \right) \right)-b \left(x-\frac{x^3}{6}+O(x^4) \right)$$ so: $$\frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=\frac{x(1+a-b)+x^3 \left(-\frac{a}{2}+\frac{b}{6} \right)}{x^3} \left(\frac{x}{f(x)} \right)^3+O(x)$$ so to have $1$ as a limit you must have: $$1+a-b=0$$ $$-a/2+b/6=1$$ which give you the answer.

Answer 4

HINT

Note that

• $\cos x=1-\frac{x^2}2+o(x^2)$
• $\sin x = x -\frac{x^3}6+o(x^3)$

then

$$\frac {x(1+a\cos x) -b\sin x}{(f(x))^3}=\frac {x+ax-a\frac{x^3}2 -bx +b\frac{x^3}6+o(x^3)}{(f(x))^3}$$