# Limits

#### Elize

I can solve this question using derivatives but how can I solve it without using derivatives?

#### v8archie

Math Team
Try writing $y = \pi - x$ and finding the limit as $y \to 0$. The limit \begin{align}\lim_{t \to 0} \frac{\cos t - 1}{t^2} &= \lim_{u \to 0} \frac{\cos 2u - 1}{(2u)^2} \\ &= \lim_{u \to 0} \frac{(1 - 2\sin^2 u) - 1}{4u^2} \\ &= \lim_{u \to 0} \frac{-2\sin^2 u}{4u^2} \\ &= -\frac12 \lim_{u \to 0} \frac{\sin^2 u}{u^2} \\ &= -\frac12 \lim_{u \to 0} \left(\frac{\sin u}{u}\right)^2 \\ &= -\frac12 \left(\lim_{u \to 0} \frac{\sin u}{u}\right)^2 \end{align}

Last edited by a moderator:
4 people

#### v8archie

Math Team
Perhaps a mod could correct the sign error: $$\lim_{u \to 0}\frac{-2\sin^2 u}{4u^2}$$

#### greg1313

Forum Staff
The limit, as posted by the OP, is $\frac12$.

Similar Math Discussions Math Forum Date
Calculus
Calculus
Real Analysis
Algebra