# Difficulty finding these two limits

1. Sep 1, 2013

### mileena

1. The problem statement, all variables and given/known data

Given that

lim f(x) = 4
x → 2

lim g(x) = -2
x → 2

lim h(x) = 0
x → 2

find the limits that exist for the problems below. If the limits do not exist, explain why.

c)
lim √f(x)
x → 2

e)
lim (g(x)/h(x))
x → 2

2. Relevant equations

lim √f(x)
x → a

=

√(lim f(x))
x → a

and

lim (f(x)/g(x))
x → a

=

(lim f(x))
x → a
/
(lim g(x))
x → a

3. The attempt at a solution

c) Since

lim √f(x)
x → 2

= 4,

√(lim f(x))
x → 2

= √4 = +2 and -2

Therefore, since 2 ≠ -2, the limit does not exist, since I don't think there can be two limits for one number.

e) Since

lim (f(x)/g(x))
x → a

=

(lim f(x))
x → a
/
(lim g(x));
x → a

lim (g(x)/h(x))
x → 2

=

(lim g(x))
x → 2
/
(lim h(x));
x → 2

= -2/0 = Undefined or ∞

The limit might equal +∞ or -∞ here, but since I don't actually have the function or the graph of the function to determine if the limit is +∞ or -∞, I will say the limit does not exist because division by 0 is not defined.

Last edited: Sep 1, 2013
2. Sep 1, 2013

### Zondrina

Your answer for $lim_{x→2} \sqrt{f(x)}$ is almost correct, but you only want to take the positive answer. Think about what is happening as you approach x=2 geometrically for some arbitrary function ( Like f(x) = 2x or f(x) = x2 for example ).

$lim_{x→2} \frac{f(x)}{g(x)}$ is incorrect. I'm getting -2.

3. Sep 1, 2013

### Enigman

√ (lim f(x))
x → 2

= √4 = +2 and -2

Therefore, since 2 ≠ -2, the limit does not exist.
-----------------------------------------------
Why? Lets take an example: $f(x)=x^2$
$√ (_{x\rightarrow 2} f(x))=\sqrt{x^2}=\left| x \right|$
so are you contending that $\left|x\right|$ is discontinuous?
EDIT: we seem to have crossed posts Z...you have solved for f/g but the OP has solved it for f/h
-seems to be a typo in the question...

4. Sep 1, 2013

### Office_Shredder

Staff Emeritus
Somehow f(x)/g(x) turned into g(x)/h(x) in the last problem.

For the square root, the square root is defined as a function as being the positive square root. It's true that if you want to solve x2 = m for some number m you have
$$x = \pm \sqrt{m}$$
but the plus/minus is there specifically because the square root of m is only returning a positive number

5. Sep 1, 2013

### mileena

Ok, thank you all three of you for answering!

First, as enigman pointed out, I made a typo in question (e). It should have been

e)

lim (g(x)/h(x))
x→2

I went back and edited my first post to correct this!

Second, thank you all three of you again for pointing out I could use an example, like f(x) = x2.

Obviously, both sides of the f(x) value are positive, so the answer is 2! I just simply wasn't thinking enough.

For problem (e), I get an answer of -2/0, and I don't know what to do from here. Normally, I would go on both sides of the x value (2 here) and see if the overall function values are positive or negative, since division by 0 is ∞, but I don't have a graph to look at or even know what the function is.

6. Sep 1, 2013

### mileena

I have an idea (which I sheepishly got from the people here who replied!): for (e), make up a function for g(x) and h(x) to test if ∞ is positive or negative!

so, g(x) = -x

and

h(x) = 2-x

so

g(x)/h(x) = -x/(2-x)

Going on both sides of 2, using 1 and 3 as examples, and plugging them into:

-x/(2-x)

we see the results on both sides are negative, so the answer is -∞!

7. Sep 1, 2013

### Enigman

For e) its undefined---http://en.wikipedia.org/wiki/Division_by_zero
see the calculus part.
:)
EDIT: crossed posts once again....
Go from left to right ie x>2 the ans. is -∞
Right to left ie x<2 the ans. is ∞

The example method was just for explanation don't use it for proofs. ;)

8. Sep 1, 2013

### Zondrina

When you get something undefined like that, it means the limit does not exist.

EDIT : You cannot simply substitute specific functions in to test the limit.

9. Sep 1, 2013

### mileena

I think what you're saying is a function must return only one value, otherwise, it's not a function, but instead is a relation.

So if f(x) = √x and x = 4,

f(x) must only be 2, not -2, since otherwise, the function would not pass the vertical line test.

10. Sep 1, 2013

### mileena

Thanks Enigman!

If a limit does not exist, we were told to write "DNE" by our professor, and give a reason, such as

lim f(x)
x→a+

lim f(x)
x→a-

I am not sure if "undefined" is allowed as an answer for us.

11. Sep 1, 2013

### Enigman

Not because of vertical line test but rather the square root operator is defined such that it returns positive values.
EDIT: Argh, this crossing of posts... I would copy the wiki line about it as a reason...

12. Sep 1, 2013

### mileena

Thanks Zondrina. I appreciate your help!

In class, we've had limits where the function had division by zero, and we had to go to one side of x to determine if the limit was +∞ or -∞.

Here is an example:

lim (-5/(x-2))
x→2+

So the limit is -∞, according to the professor.

Of course, you are right, but I am just trying to see where I went astray by using the above example from class.

13. Sep 1, 2013

### Enigman

Go about it other way
lim (-5/(x-2)
x→2- and you will get +∞ as the answer ; hence limit doesn't exist.

14. Sep 1, 2013

### mileena

Ok, somehow, I never learned about the square root operator only returning positive values back in high school, but I think I did read about somewhere in my independent study.

I am going to the wikipedia link you posted right now!

15. Sep 1, 2013

### mileena

That's a good point Enigman. But in class we were dealing with one-sided limits, so as x approaches 2 from the right (or positive) side, we got -∞ for just that limit. I am not sure if if the limits as you approach from both the left and right sides actually agree in value, then that is the limit for x, or if the answer is "does not exist".

16. Sep 1, 2013

### Enigman

Its not the limit for x but the limit of the function.
Limit of function is said not to exist at x=x' when the left-hand limit, the right-hand limit and f(x') are not all equal.

17. Sep 1, 2013

### mileena

Thanks Enigman! I am learning, little by little. This is one of my more difficult classes, but because of the homework for each class session, and the short quiz in each class, I am forced to learn, which a good thing. It helps that the professor is really a good teacher too, and that I have this board to help me out if needed.

Someday, I hope to be posting here, as a mentor too!

18. Sep 1, 2013

### Enigman

Its a race then :D

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