[SOLVED] The Squeeze Theorem

**Mike9182** · August 7th 2009, 07:26 AM

How do you solve this?

$\lim_{x\to\frac{\pi}{3}} (\sqrt{3} \sin x -2x)$

**VonNemo19** · August 7th 2009, 07:51 AM

Originally Posted by Mike9182

How do you solve this?

$\lim_{x\to\frac{\pi}{3}} (\sqrt{3} \sin x -2x)$

Direct Substitution...

That's all. There's no need for any advanced techniques here.

I'll do it... Here's a step by step of all the relevant theorems of limits that apply.
1. The limit of a sum is the sum of the limits.

$\lim_{x\to\frac{\pi}{3}}(\sqrt{3}\sin x-2x)=\lim_{x\to\frac{\pi}{3}}\sqrt{3}\sin{x}-\lim_{x\to\frac{\pi}{3}}2x$

2. The limit of a product is the product of the limits.

$=\lim_{x\to\frac{\pi}{3}}\sqrt{3}\cdot\lim_{x\to\f rac{\pi}{3}}\sin{x}-\lim_{x\to\frac{\pi}{3}}2\cdot\lim_{x\to\frac{\pi} {3}}x$

4. The limit of a constant is that constant

$=\sqrt{3}\lim_{x\to\frac{\pi}{3}}\sin{x}-2\lim_{x\to\frac{\pi}{3}}x$

5. The ovious limit

$=\sqrt{3}\lim_{x\to\frac{\pi}{3}}\sin{x}-2\frac{\pi}{3}$

6. The limit of transcendentals

$=\sqrt{3}\frac{\sqrt{3}}{2}-2\frac{\pi}{3}=\frac{3}{2}-\frac{2\pi}{3}$

You can see that there is no need for the squeeze theorem.

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