Trigonometry Limits: Solving lim x -> 0 sin x / sin(x/2)

[Deathfish]

Homework Statement

Find lim x -> 0 [tex]\frac{sin x}{sin\frac{x}{2}}[/tex]

The Attempt at a Solution

Since period of sin 2(x/2) is T/2 compared to period T of sin (x/2)

sin 2(x/2) nears twice the value of sin (x/2) for all values of x approaching zero.

Therefore lim x -> 0 [tex]\frac{sin x}{sin\frac{x}{2}}[/tex] = 2

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somehow i find the reasoning flawed, anyone can offer a better solution?

 

[magicarpet512]

Did you think to try L'Hospital's Rule?

 

[HallsofIvy]

if you don't want to use Calculus (though this is in the Calculus section!), use the double angle formula, sin(2a)= 2 sin(a)cos(a) to write this as
[tex]\frac{2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}{sin\left(\frac{x}{2}\right)}[/tex]