Limit of sin(x)/x: Converting to sin(1/x)

[azatkgz]

Homework Statement

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

(i)$$\lim_{x\rightarrow\infty}\frac{sinx}{x}$$

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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[azatkgz]

Hurkyl?

 

[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

 

[azatkgz]

But from sqeeze theorem we get 0.

 

[JasonRox]

But from sqeeze theorem we get 0.

Yeah, and what's the problem?

 

[azatkgz]

Ok ,then.

 

[azatkgz]

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

 

[Kreizhn]

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