Limit of a trigonometric function (Advanced problem)

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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 [tex] \underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \frac{x}{2}-1}{x\sin x} [/tex]

Homework Equations





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.

Thank you for your help!

I) Pure algebraic manipulation:

[tex] \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} [/tex]

II) Manipulation with conjugate method:

[tex] \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} [/tex]
 
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