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Limit of a trigonometric function (Involved problem)

  • Thread starter vertciel
  • Start date
  • #1
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Homework Statement



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

[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]
 

Answers and Replies

  • #2
737
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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.
 
  • #3
867
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First let u=x/2, then x=2u. As x/2→0, u→0 also.


[tex]\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)}[/tex]

The numerator becomes tan2u, then see where you can go from there.
 
  • #4
63
0
Thank you very much for your replies, Tedjn and Bohrok.

I was able to evaluate this limit.
 

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