Finding the limit in terms of a without using L'Hospitals

Question

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.

Answer 1

Hint: Multiply both numerator and denominator of the expression by $1+\sqrt{1+x^2}$ and your expression will have the factor $-\dfrac{1-\cos(ax)}{x^2}$. Then use the identity $1-\cos(ax) = 2\sin^2\left(\frac{ax}{2}\right)$ and the known fact about limit: $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$ to obtain the desire result.

Answer 2

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 $\dfrac{a^2}{4}$:

$\lim_\limits{x \to 0} -\dfrac{[2\sin^2(\frac{ax}{2})][1 - \sqrt{1+x^2}]}{x^2} \cdot \dfrac{\frac{a^2}{4}}{\frac{a^2}{4}} = \lim_\limits{x \to 0} \dfrac{[\sin^2(\frac{ax}{2})][-\frac{1}{2}a^2][1 - \sqrt{1+x^2}]}{\dfrac{a^2x^2}{4}}$

$=\lim_\limits{x \to 0} \dfrac{\sin^2(\frac{ax}{2})}{\dfrac{a^2x^2}{4}} \cdot \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})$ $= \left( \lim_\limits{x \to 0} \dfrac{\sin(\frac{ax}{2})}{\dfrac{ax}{2}}\right)^2 \cdot \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})$

$= \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})=-a^2$