Evaluate $\lim_{x \to 0} \left( \frac{a \cos x}{a+ b\sin x} \right)^{1/x}$, where is the mistake?

Question

I am not interested in the answer. I am interested in where I have made an error. I am to evaluate:

$$\lim_{x \to 0} \left( \frac{a\cos x}{a+b\sin x} \right)^{1/x}$$

First, simplify

$$\lim_{x \to 0} \left( \frac{\cos x}{1 + \frac{b}{a} \sin x} \right)^{1/x}$$

Now, multiply top and bottom by $-1$, and add and deduct $1$ from the top:

$$\lim_{x \to 0} \left( \frac{(1-\cos x) + 1}{-(1+\frac{b}{a}\sin x)} \right)^{1/x}$$

Now:

$$\lim_{x \to 0} \left( \frac{1-\cos x}{-(1+\frac{b}{a}\sin x)} \right)-\lim_{x \to 0}\frac{1}{1+\frac{b}{a}\sin x} = \\ -\lim_{x \to 0} \left(\left( \frac{1-\cos x}{x^2} \right)\left( \frac{x^2}{1+\frac{b}{a}\sin x}\right)\right) -1 \\ -\left(\frac{1}{2} \cdot 0 \right)-1 = -1$$

Which is not correct...

Answer 1

Hint

Let $$f(x)=\frac{a\cos(x)}{a+b\sin(x)}$$

we have $$\lim_0f(x)=1$$ thus

$$\lim_0\frac{\ln(f(x))}{f(x)-1}=1.$$

write it as

$$(f(x))^\frac 1x=e^{\color{red}{\frac 1x} \ln(f(x))}$$

but

$$\frac{f(x)-1}{x}=\frac{a(\cos(x)-1)-b\sin(x)}{x(a+b\sin(x))}$$ As you know, goes to $\frac{-b}{a}$

the limit is

$$e^{\frac{-b}{a}}$$