# 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)
[

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 $$cos(x)$$?