Limit $\lim_{x\to 0}\frac{\sin(ax)}{\sqrt{bx+c^2}-c}$

Question

How to find the following parametric limit without l' hospital rule? $$\lim_{x\to 0}\frac{\sin(ax)}{\sqrt{bx+c^2}-c}$$ with $$b>0$$

Answer 1

Hint:

For $x\neq 0$ we can put

\begin{align*} \frac{\sin (ax)}{\sqrt{bx+c^2}-c}&=\frac{\sin (ax)}{\sqrt{bx+c^2}-c}\cdot\frac{{\sqrt{bx+c^2}+c}}{\sqrt{bx+c^2}+c}\\ &=\frac{\sin (ax)}{bx}\cdot \left(\sqrt{bx+c^2}+c\right) \end{align*}

Answer 2

Taylor series or equivalents are useful.

Consider first the denominator (assuming $c >0$) $$\sqrt{c^2+bx}-c=c \left(\sqrt{1+\frac b {c^2}x}-1 \right)$$ Now, use the generalized binomial theorem or Taylor series to get $$\sqrt{1+\frac b {c^2}x}=1+\frac{b }{2 c^2}x+O\left(x^2\right)$$ and $$\sin(ax)=ax+O\left(x^2\right)$$

I am sure that you can take it from here.