Limit of a trigonometric function (Involved problem)

[vertciel]

Homework Statement

Evaluate $$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \frac{x}{2}-1}{x\sin x}$$, WITHOUT using l'Hôpital's rule.

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:

$$\begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \tfrac{x}{2}-1}{x\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos \tfrac{x}{2}}{\cos \tfrac{x}{2}}\div \frac{1}{x\sin x} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos \tfrac{x}{2}}{x\sin x\cos \tfrac{x}{2}} \right) \\ & =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1-\cos \tfrac{x}{2}}{\tfrac{x}{2}} \right)\times \underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\tfrac{1}{2}}{\sin x\cos \tfrac{x}{2}} \right) \\ \end{align}$$

II) Manipulation with conjugate method:

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

 

[Tedjn]

One way to approach this problem is to transform everything into sine. Then, you can make use of the sin(x)/x limit. To transform the difference 1 - cos(x/2) into a sine, for example, substitute cos(0) for 1 and make use of the sum to product formulas, or recognize it as part of a power reducing formula for sine.

 

[Bohrok]

First let u=x/2, then x=2u. As x/2→0, u→0 also.


$$\lim_{x\rightarrow 0}\frac{\sec\frac{x}{2} -1}{x\sin x} = \lim_{u\rightarrow 0}\frac{\sec u - 1}{2u\sin2u} \times \frac{\sec u + 1}{\sec u + 1} = \lim_{u\rightarrow 0}\frac{\sec^2u - 1}{2u\times 2\sin u \cos u(\sec u + 1)}$$

The numerator becomes tan²u, then see where you can go from there.

 

[vertciel]

Thank you very much for your replies, Tedjn and Bohrok.

I was able to evaluate this limit.