# Limit of a trigonometric function (Advanced problem)

vertciel
Adminstrator: This is a double post, so please feel free to delete this one.

The relevant post is "Limit of a Trigonometric Function (Involved Problem)".

## Homework Statement

Evaluate $$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \frac{x}{2}-1}{x\sin x}$$

## The Attempt at a Solution

Hello there,

I tried to evaluate this limit using two different approaches, both of which still leave the limit in indeterminate form when 0 is substituted.

\begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \tfrac{x}{2}-1}{x\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos \tfrac{x}{2}}{\cos \tfrac{x}{2}}\div \frac{1}{x\sin x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos \tfrac{x}{2}}{x\sin x\cos \tfrac{x}{2}} \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos \tfrac{x}{2}}{\tfrac{x}{2}} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\tfrac{1}{2}}{\sin x\cos \tfrac{x}{2}} \right) \\ \end{align}
\begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \tfrac{x}{2}-1}{x\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos \tfrac{x}{2}}{x\sin x\cos \tfrac{x}{2}} \right) \\ & \text{Let u = }\tfrac{x}{2}.\text{ Since }\underset{u\to 0}{\mathop{\lim }}\,u=0\text{, u still tends to 0}\text{.} \\ & \underset{u\to 0}{\mathop{\lim }}\,\frac{1-\cos u}{\tfrac{u}{2}\cos u\sin \tfrac{u}{2}}\left( \frac{1+\cos u}{1+\cos u} \right)=\underset{u\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{2}}u}{\tfrac{u}{2}\cos u\sin \tfrac{u}{2}(1+\cos u)} \\ & =\underset{u\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}u}{\tfrac{u}{2}\cos u\sin \tfrac{u}{2}(1+\cos u)} \\ \end{align}