Taking a limit of a function where it does not exist?

[Zerius]

Homework Statement

lim sqrt(x-9)-3
x->0 sqrt(x)

Homework Equations

l'hop?

The Attempt at a Solution

as far as i got was conjugate the top and arriving somewhere where it is still undefined. =(

 

[TD]

Are you sure it isn't sqrt(x+9) instead of sqrt(x-9)? That would make more sense.

 

[franznietzsche]

Homework Statement

lim sqrt(x-9)-3
x->0 sqrt(x)

Homework Equations

l'hop?

The Attempt at a Solution

as far as i got was conjugate the top and arriving somewhere where it is still undefined. =(

$$\frac{d}{d x} (\sqrt{x-9}-3) = ?$$
and
$$\frac{d}{d x} \sqrt{x} = ?$$
(you get to do these)
Try L'hospital. You can't try to conjugate the top with these square roots flying around. Also, +9 would make more sense, but doesn't change your method of solution.

 

[Zerius]

Thanks, it was actually x+9 as you suggested. much easier to solve in this case, I conjugated the top which brings the Top to "x" and the bottom "sqrt(x)(sqrt(x+9)+3)" division of the variable brings

$$\frac{\sqrt{x}}{\sqrt{x+9}+3}$$

which makes lim x-> 0 go to 0/6 = 0

but I am still curious, is it possible to take a limit of a function where the function does not exist such as in this case?

Thanks

 

[TD]

If it was like you wrote (so sqrt(x-9)), then it doesn't make any sense to take the limit for x to zero since sqrt(x-9) isn't defined for x<9. This is why usually in the formal definition of a limit, the 'a' of the limit "for x going to a" has to be a limit point of the domain of your function. Notice that the function doesn't have to be defined in a itself, but this guarantees that f exists in a neighbourhood arround a.

 

[JG89]

If you mean what I think you mean, it's possible. For example, as x tends to 0, the function $$\frac{1}{x^2}$$ goes to infinity.

 

[TD]

Because 1/x² exists in a neighbourhood arround x=0, that's not the case for e.g. sqrt(x-9).

 

[franznietzsche]

Because 1/x² exists in a neighbourhood arround x=0, that's not the case for e.g. sqrt(x-9).

You can have complex numbers as answers. I just wouldn't expect it in a first course in differential calculus.

 

[JG89]

TD, you're right. Strictly speaking about the set of real numbers, should the OP's question really be asking for a right-handed limit?

 

[TD]

You can have complex numbers as answers. I just wouldn't expect it in a first course in differential calculus.

I'm assuming the OP is studying limits of real-valued functions with a subset of the reals as domain.

TD, you're right. Strictly speaking about the set of real numbers, should the OP's question really be asking for a right-handed limit?

The original question (with sqrt(x-9) and x to zero) makes no sense, not even when you're talking about left- or right-handed limits. But if you're looking at sqrt(x), which is not defined for x<0, then you could say you're dealing with a right-handed limit. It really depends on how (formal) you defined the limit. It's certainly possible to talk about "the limit" in such a case, if you're restricting the values you're looking at to the intersection of a neighbourhood arround x=a and the domain of the function (without a itself). Then, in a case like sqrt(x), the definition already "tells you" you're only looking at x>0. I'm not sure if I'm being very clear

 

[JG89]

I get what you mean. And yes, I was speaking about x + 9 in the square root sign.

 

[TD]

Well in the case of sqrt(x+9) and taking the limit for x to 0, there's no problem with coming from either left or right. You (might) have that problem when you're taking the limit for x to -9, which would only be possible from the right side (which is the same situation as sqrt(x) for x to 0, of course ).

 

[JG89]

Sorry, I didn't make myself clear. I know that the original question should be $$\frac{\sqrt(x+9)-3}{\sqrt(x)}$$. But you must take this limit from the right because if you took it from the left, you would have the square root of a negative number in the denominator. Am I right?

 

[TD]

Right, I wasn't thinking about the sqrt(x) in the denominator anymore. So you can certainly take the right-handed limit, but considering what I said (or was trying to say...) in post #10, it's also possible to talk about "the limit" for x to 0.

 

[Zerius]

Thanks everyone, that was in preparation for my final today, which I hope I did not fail. Oh well! This is from a first year single variable differential calc course @ UBC.