$\lim_{x \to 0} \frac{\cos(\sqrt{a+x})-\cos(\sqrt{a})}{x}$

Question

$$\lim_{x \to 0} \frac{\cos(\sqrt{a+x})-\cos(\sqrt{a})}{x}$$

The answer to this question is $\frac{d}{da}\cos(\sqrt{a})$
i.e., $\begin{eqnarray*}-\frac{\sin(\sqrt{a})}{2\sqrt{a}}\end{eqnarray*}$
but I want to get answer by using general rules of limits. Also I don't want to use L'Hospital's rule

Answer 1

We have $$y=\lim_{x\to0}\frac{\cos\sqrt{x+a} -\cos\sqrt a}{x} = \lim_{x\to0}-\frac{2}{x}\sin\left(\frac{\sqrt{x+a}-\sqrt a}{2}\right)\sin\left(\frac{\sqrt{x+a}+\sqrt a}{2}\right)$$

$$y=-2\lim_{x\to0}\frac{\sin\left(\frac{\sqrt{x+a}-\sqrt a}{2}\right)}{x+a-a}\cdot\lim_{x\to0}\sin\left(\frac{\sqrt{x+a}+\sqrt a}{2}\right)$$

Now, $ x=x+a-a = (\sqrt{x+a}-\sqrt a)(\sqrt{x+a}+\sqrt a) \\x\to0\implies \sqrt{x+a}-\sqrt a \to0 $

Let $z=\sqrt{x+a}-\sqrt a $

So, $$y=-\lim_{z\to0}\frac{\sin(\frac{z}{2})}{z/2}\cdot\lim_{x\to0}\frac{1}{\sqrt{x+a}+\sqrt a}\cdot\lim_{x\to0}\sin\left(\frac{\sqrt{x+a}+\sqrt a}{2}\right)$$

$$y = -1\cdot\frac{1}{2\sqrt a}\cdot\sin\frac{2\sqrt a}2 = -\frac{1}{2\sqrt a}\sin(\sqrt a)$$

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Formulas used

$\bullet \ \cos A- \cos B = -2\sin\frac{A-B}{2}\sin\frac{A+B}{2}$
$\bullet \ \lim_{x\to0}\frac{\sin x}{x} = 1$