Homework Help: Simple Limit.

1. Oct 18, 2007

azatkgz

1. The problem statement, all variables and given/known data

Find the following Limit in terms of the number
$$\alpha=\lim_{x\rightarrow 0}\frac{sinx}{x}$$

(i)$$\lim_{x\rightarrow\infty}\frac{sinx}{x}$$
3. The attempt at a solution

$$\alpha=\lim_{x\rightarrow\infty}\frac{sin(1/x)}{1/x}$$
But I don't know how to convert sin(x) to sin(1/x)
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2. Oct 18, 2007

azatkgz

Hurkyl?

3. Oct 18, 2007

Kreizhn

I'm not terribly sure how you got

$$\alpha=\lim_{x\rightarrow\infty}\frac{sin(1/x)}{1/x}$$

Note that you cannot simply divide through by x when dealing with trigonometric functions. In this case you'll want to use the Squeeze Theorem.

Note that $$|sin x| \leq 1$$

This implies that $$\left| \frac{sin x}{x} \right| \leq \frac{1}{x}$$

Now you can apply the Squeeze Theorem

4. Oct 18, 2007

azatkgz

But from sqeeze theorem we get 0.

5. Oct 18, 2007

JasonRox

Yeah, and what's the problem?

6. Oct 18, 2007

azatkgz

Ok ,then.

7. Oct 18, 2007

azatkgz

Can we in the similar manner say that $$\lim_{x\rightarrow\infty}\frac{cosx}{x}=0$$

8. Oct 18, 2007

Kreizhn

Is there any point in the argument which doesn't hold as a result of switching $$sin(x)$$ to $$cos(x)$$?