Why is this limit not well defined?

[adelaide87]

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


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

 

[Tedjn]

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

 

[adelaide87]

Left.

Does it help to put the cos on the bottom?

 

[Dickfore]

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

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

 

[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?

 

[adelaide87]

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

 

[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?

 

[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.

 

[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?

 

[Tedjn]

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?

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.

 

[adelaide87]

Ah wait a sec, anything greater than x=1 is undefined. But you don't have to even worry about anything greater than 1, because its x > -infinity. Thats why i don't see how it is undefined.

 

[Dickfore]

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

 

[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?

 

[Dickfore]

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?

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

 

[adelaide87]

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


Should the signs be less than/equal to?

 

[Dickfore]

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


Should the signs be less than/equal to?

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 $a/b \ge 0$?

 

[Dickfore]

I will not answer anymore in this thread.