# Finding the Limit of sin|x|/x

• math2010
In summary, the conversation is about finding the limit of a function, specifically \lim_{x\to 0}\frac{sin|x|}{x}. The person asking the question knows that the limit does not exist, but is unsure of how to arrive at that answer. They mention knowing the limit of \lim_{x\to 0}\frac{sinx}{x}, but are unsure of how that applies to this situation. Another person suggests looking at the two one-sided limits, and the first person is confused about what those are. The second person explains that for the first one-sided limit, |x| = x, and for the second one-sided limit, |x| = -x.

## Homework Statement

I want to find the limit:

$$\lim_{x\to 0}\frac{sin|x|}{x}$$

## The Attempt at a Solution

I know that the answer must be "limit doesn't exist" but I don't know how to arrive at that answer. I know that $$\lim_{x\to 0}\frac{sinx}{x}=1$$ but apparently it's a very different situation. Can anyone show me how to find this limit?

Look at the two one-sided limits, and see if they are the same or different.

Mark44 said:
Look at the two one-sided limits, and see if they are the same or different.

What are the two one sided limits? That's what I don't get!

lim x -->0+
lim x -->0-

For the first, |x| = x
For the second, |x| = -x

## What is the limit of sin|x|/x as x approaches 0?

The limit of sin|x|/x as x approaches 0 is equal to 1. This can be shown by using the squeeze theorem and the fact that the absolute value of sin|x| is always less than or equal to 1.

## Why is it important to find the limit of sin|x|/x?

Finding the limit of sin|x|/x is important because it helps us understand the behavior of the function as x approaches 0. This can be useful in solving problems involving rates of change and optimization.

## What is the graph of sin|x|/x?

The graph of sin|x|/x is a continuous function with a removable discontinuity at x=0. It has a horizontal asymptote at y=1 and the graph oscillates between -1 and 1 as x approaches 0 from either side.

## Can the limit of sin|x|/x be evaluated using L'Hopital's rule?

Yes, the limit of sin|x|/x can be evaluated using L'Hopital's rule. By taking the derivatives of both the numerator and denominator, we can rewrite the limit as the limit of cos|x|/1, which is equal to 1 as x approaches 0.

## How does the limit of sin|x|/x compare to the limit of cos|x|/x as x approaches 0?

The limit of sin|x|/x is equal to 1, while the limit of cos|x|/x is undefined. This is because the function cos|x|/x has a vertical asymptote at x=0, while the function sin|x|/x does not.