TITLE: Finding the limit in terms of a without using L'Hospitals
QUESTION [3 upvotes]: Find $\;\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}\;$ in terms of a without using L'Hospitals Rule.
I first graphed the function and noticed that that limit tends to go towards $-a^2$
I also tried this approach:
$\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}=\lim_\limits{x \to 0} \dfrac{2 \cdot \frac{1-\cos(ax)}{2}}{1-\sqrt{1+x^2}} = \lim_\limits{x \to 0} \dfrac{2\sin^2(\frac{ax}{2})}{1 - \sqrt{1+x^2}} $
However, I'm not sure how to proceed. I tried rationalizing the denominator but it only makes it more tangled.

REPLY [1 votes]: Thank to @Wang YeFei for the hint.
$\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}=\lim_\limits{x \to 0} \dfrac{2 \cdot \frac{1-\cos(ax)}{2}}{1-\sqrt{1+x^2}} = \lim_\limits{x \to 0} \dfrac{2\sin^2(\frac{ax}{2})}{1 - \sqrt{1+x^2}} $
Then we proceed to multiply the numerator and denominator by $1 - \sqrt{1+x^2}$:
$\lim_\limits{x \to 0} \dfrac{2\sin^2(\frac{ax}{2})}{1-\sqrt{1+x^2}} \cdot \dfrac{1 - \sqrt{1+x^2}}{1 - \sqrt{1+x^2}} = \lim_\limits{x \to 0} -\dfrac{[2\sin^2(\frac{ax}{2})][1 - \sqrt{1+x^2}]}{x^2}$
Then, to be able to use the fact that $\lim_ \limits{x \to 0} \dfrac{\sin x}{x}=1$, we multiply the numerator and denominator by $\dfrac{a^2}{4}$:
$\lim_\limits{x \to 0} -\dfrac{[2\sin^2(\frac{ax}{2})][1 - \sqrt{1+x^2}]}{x^2} \cdot \dfrac{\frac{a^2}{4}}{\frac{a^2}{4}} = \lim_\limits{x \to 0} \dfrac{[\sin^2(\frac{ax}{2})][-\frac{1}{2}a^2][1 - \sqrt{1+x^2}]}{\dfrac{a^2x^2}{4}}$
$=\lim_\limits{x \to 0} \dfrac{\sin^2(\frac{ax}{2})}{\dfrac{a^2x^2}{4}} \cdot \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})$
$= \left( \lim_\limits{x \to 0} \dfrac{\sin(\frac{ax}{2})}{\dfrac{ax}{2}}\right)^2 \cdot \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})$
$= \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})=-a^2$
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