# Trig Limit question

1. Jul 5, 2009

### helpcalc

Could someone help me with this one. My teacher says I'm wrong when I solved this limit as undefined. thanks!

lim as x goes to zero of (sin x tan x -1)/4x

2. Jul 6, 2009

### Bohrok

If you mean (sin(x)tan(x)-1)/(4x), then it would be undefined. I have a feeling that the -1 is causing a problem since you may be missing parentheses that would change the problem and let a limit exist.

3. Jul 6, 2009

### n!kofeyn

$$\lim_{x\to 0} \frac{\sin x \tan x - 1}{4x}$$
then the limit is undefined, and your teacher is wrong. The way to determine this is to look at the limit as x approaches 0 from the left and then from the right.

The limit of the numerator as x approaches 0 is -1. As x approaches 0 from the left, x is negative and so the limit will be positive infinity. As x approaches 0 from the right, x is positive, so you will get negative infinity. Therefore, the limit does not exist since the left and right handed limits aren't equal.

4. Jul 6, 2009

### helpcalc

thanks for the responses but when i graph this function at 0 the function is zero. It looks like it is continuous and goes to zero from the left and right but i can't find a mathematical way to solve it.

5. Jul 6, 2009

### Cyosis

Well you must have drawn the graph incorrectly. For example what value did you get for x=0.1 and x=-0.1?

6. Jul 6, 2009

### helpcalc

for -.01 the limit is .0025 and for .01 the limit is -.0025. looking at a graphing calculator using the table function, it looks like it converges to 0.

7. Jul 6, 2009

### Cyosis

I see, that means we're talking about different functions and so are the other posters in this thread.

The limits you describe correspond with the function:

$$x \left(\frac{\sin x \tan x -1}{4}\right)$$

We assumed you meant:

$$\frac{\sin x \tan x - 1}{4x}$$

Could you clarify which one it is?

8. Jul 6, 2009

### helpcalc

my mistake on the calculator. it is definitely the second one. looking at the graph my teacher must be wrong (I hope!!!!!!) thanks for your help!!!!!!!!!

9. Jul 6, 2009

### Cyosis

If it's the second one then the limit does not exist.