# Limit of Impossibility

## 12 December 2013

This is a really cool trick I only recently learned. Evaluating limits has useful applications, and this little trick allows the evaluation of limits which, on the surface, seem impossible. Consider

$\lim_{x\to 0}\frac{2\cos 2x-2+4x^2}{2x^4}.$The denominator poses a problem, as $\lim_{x\to 0}2x^4 = 0$, therefore
the entire limit is of the form $\frac{-2}{0}$, which suffice it to
say is not acceptable. It isn’t one of the forms to which L’Hôpital’s
rule^{1} applies.

We expect the limit to be evaluable, however, by inspecting the graph.

However, there is another way to evaluate this limit, though it involves making it appear very complex. By expanding $\cos 2x$ using the Maclaurin series for $\cos x$, the limit can be expanded.

$\begin{array}{rl} & \lim_{x\to 0}\frac{ 2\sum_{k=0}^{\infty}\frac{(-1)^k(2x)^{2k}}{(2k)!}-2+4x^2 }{2x^4} \\ =& \lim_{x\to 0}\frac{ 4x^2-2+ 2\left[ 1- \frac{(2x)^2}{2!}+ \frac{(2x)^4}{4!}- \frac{(2x)^6}{6!}+ \cdots \right]}{2x^4} \\ =& \lim_{x\to 0}\frac{ 4x^2 -2+ 2- \frac{2(2x)^2}{2!}+ \frac{2(2x)^4}{4!}- \frac{2(2x)^6}{6!}+ \cdots }{2x^4} \\ =& \lim_{x\to 0}\frac{ \frac{2(2x)^4}{4!}- \frac{2(2x)^6}{6!}+ \cdots }{2x^4} \\ =& \lim_{x\to 0}\frac{ 2x^4\cdot\left[ \frac{2^4}{4!}- \frac{2^6x^2}{6!}+ \frac{2^8x^4}{8!}- \cdots \right]}{2x^4} \\ =& \frac{16}{24}=\frac{2}{3} \\ \end{array}$The remaining terms from the Maclaurin series are eliminated because they tend to zero when the limit is evaluated, which leaves us with the expected value of $\frac{2}{3}$. That is unqualifiedly cool.

*This is a solution for §9.4 Exercise #9 from* Calculus: Early Transcendentals.