Limit of x^2/(1 - cos x) at 0 Is 2

Turn the known limit over

The quotient law allows this step because the limit being inverted is $\frac{1}{2}$, and $\frac{1}{2} \neq 0$. State that condition when you use it, since it is the only thing standing between this move and a false one.

Why the nonzero condition earns its place

Take the same numerator over $x$ instead of $x^{2}$. That limit is $0$, so its reciprocal has nothing finite to be.

Flipping a limit of $0$ never gives a finite value. Here $\frac{1-\cos x}{x}$ changes sign at $0$, so its reciprocal has no limit at all, not even an infinite one. When the vanishing quantity keeps one sign, as $1-\cos x$ itself does, the reciprocal runs to $+\infty$ instead. Flipping a nonzero limit is safe.

A half-angle check

The bracket tends to $1$ by the sine limit, so the $2$ out front is the answer. Two independent routes landing on $2$ is worth more than one route repeated.

The mistakes students make

The first is by far the most common, and it is worth a deliberate check that your quotient is the right way up.

• Giving $\frac{1}{2}$, which is the standard limit as it is usually stated, and missing that this quotient is the other way up.
• Treating $1 - \cos x$ as roughly $x$, the way $\sin x$ is. That turns the quotient into $x$ and gives $0$. The correct size is $\frac{x^{2}}{2}$.
• Calling it undefined because the denominator vanishes at $0$. The numerator vanishes too, so the form is $\frac{0}{0}$ and there is work to do.