# Why is this limit not well defined?

1. Jun 21, 2010

### adelaide87

Lim
x--> -Infinity cos^-1[(3-x)/(x+4)]

Book says this limit is not well defined, why is that?

2. Jun 21, 2010

### Tedjn

As x approaches -∞, (3-x)/(x+4) approaches -1. From which direction?

3. Jun 21, 2010

### adelaide87

Left.

Does it help to put the cos on the bottom?

4. Jun 21, 2010

### Dickfore

$$\cos^{-1} x \equiv \arccos x$$

is the inverse function of the cosine, not $$1/\cos x$$.

5. Jun 22, 2010

### Dickfore

Find the domain of definition for the composite function above. Hint: $\arccos x$ is only defined for $|x| \le 1$. Is the limit point an element of this domain?

6. Jun 22, 2010

### adelaide87

Im confused. I have never done a question like this one and it randomly asked it.

7. Jun 22, 2010

### Dickfore

Well, you can only learn by doing things you've never done before. If you only do things you already know how to do, there is no need of learning anything, is there?

8. Jun 22, 2010

### Tedjn

Good advice. What exactly is confusing you? You've already realized that (3-x)/(x+4) approaches -1 from the left. Try to combine that with what Dickfore posted.

9. Jun 22, 2010

### adelaide87

So is it that inverse cosine that makes it not well defined? Because as you plug -ive x values in it goes to -infinity. Correct?

10. Jun 22, 2010

### Tedjn

What did you mean by the bolded part? What goes to -∞? Of course, that there is an arccosine is important; it's the basis behind the problem.

11. Jun 22, 2010

### adelaide87

Ah wait a sec, anything greater than x=1 is undefined. But you dont have to even worry about anything greater than 1, because its x > -infinity. Thats why i dont see how it is undefined.

12. Jun 22, 2010

### Dickfore

Please review the topic on function composition and how to find domains and ranges of functions. Also, review inequalities involving absolute values.

13. Jun 22, 2010

### adelaide87

Would it be correct to say it's not well defined because f(x) is not defined for x < -4?

Because (3 - x)/(x + 4) -> -1 as x -> -infinity, its approaching from values that are less than -1. arccos[(3 - x)/(x + 4)] is only defined when -1 < (3 - x)/(x + 4) < 1.

Does that seem correct?

14. Jun 22, 2010

### Dickfore

You have the right idea now, but you made a mistake with the inequalities.

15. Jun 22, 2010

### adelaide87

For -1 < (3 - x)/(x + 4) < 1.

Should the signs be less than/equal to?

16. Jun 22, 2010

### Dickfore

Yes, but that's not the important issue. You should do one of them like this:

$$\frac{3 - x}{x + 4} \le 1$$

$$\frac{3 - x}{x + 4} - 1 \le 0$$

$$\frac{3 - x - x - 4}{x + 4} \le 0$$

$$\frac{-1 - 2x}{x + 4} \le 0 /\cdot (-1)$$

$$\frac{2x + 1}{x + 4} \ge 0$$

Notice, that, because I multiplied the whole inequality by a negative number, the sign of the inequality changed in the last step. This is very important and you should keep it in mind.

Now, question. When is $a/b \ge 0$?

17. Jun 22, 2010

### Dickfore

I will not answer anymore in this thread.

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