Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 30

Determine the $\displaystyle \lim_{x \to 0} \frac{\cos mx - \cos nx}{x^2}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to 0} \frac{\cos mx - \cos nx}{x^2} = \frac{\cos m(0) - \cos n(0)}{0^2} = \frac{1-1}{0} = \frac{0}{0} \text{ Indeterminate}$

Thus by applying L'Hospital's Rule...

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{\cos mx - \cos nx}{x^2} &= \lim_{x \to 0} \left[ \frac{(- \sin mx)(m) - (-\sin nx)(n)}{2x} \right]\\
\\
&= \lim_{x \to 0} \left[ \frac{n \sin(nx) - m \sin(mx)}{2x} \right]
\end{aligned}
\end{equation}
$


We will still get indeterminate form if we evaluate the limit, so we apply L'Hospital's Rule once more...


$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \left[ \frac{n \sin(nx) - m \sin(mx)}{2x} \right] &= \lim_{x \to 0} \left[ \frac{n \cos (nx)(n)-m\cos(mx)(m)}{2} \right]\\
\\
&= \lim_{x \to 0} \left[ \frac{n^2 \cos (nx) - m^2 - \cos (mx)}{2} \right]\\
\\
&= \frac{n^2 \cos(n(0)) -m^2 \cos(m(0))}{2}\\
\\
&= \frac{n^2(\cos 0 ) - m^2(\cos 0)}{2}\\
\\
&= \frac{n^2(1) - m^2(1)}{2}\\
\\
&= \frac{n^2-m^2}{2}
\end{aligned}
\end{equation}
$