# Solving this limit without a calculator

1. Oct 17, 2004

### Shay10825

Is there anyway to do this without calculatior and without graphing it?

lim (x-3)/(x-2)
x->2+

2. Oct 17, 2004

### Sirus

Yes. Think of it this way: in the denominator, you are subtracting 2 from a value just barely larger than 2; what does that equal? You should be able to figure it out from there.

3. Oct 17, 2004

### Shay10825

I know the answer is neg inf. But how do you know if the answer is inf, a number or if it does not exist? Like for the problem:

lim (x^2)/(x^2+16)
x->4-

The answer is .8 but how would I know this without a calc and without graphing it?

4. Oct 18, 2004

### Shay10825

Would taking the derivative help? I heard that taking the derivative could help in some cases. Is this true?

5. Oct 18, 2004

### JonF

lim (x-3)/(x-2)
x->2+

in this case, the denominator approaches 0 while the numerator approaches 1. What happens to a fraction when it’s denominator gets small?

Try comparing 1/10, 1/1, 1/.1, and 1/.00000001. If it’s denominator was infinitely close to 0 it would approach infinity no mater what finite value it’s numerator is.

lim (x^2)/(x^2+16)
x->4-

would be 1/2.

6. Oct 18, 2004

### Shay10825

How do you know this is continuous without graphing it and without a calculator?

lim (x^2)/(x^2+16)
x->4-

How do you know this is not continuous without graphing it and without a calculator? Do all functions that are not continuous have a lim of inf or neg inf?

lim (x-3)/(x-2)
x->2+

Last edited: Oct 18, 2004
7. Oct 18, 2004

### JasonRox

First you need to define continuity before you can ask what is continuous.

A continuous function satisfies the following:

lim f(x) = f(a)
x->a

Get it? A function is continuous when a is in its domain. The domain is defined as what you can "put in" the function.

Here is the function box:

DOMAIN -> F -> RANGE

F spits out Range. For every number in the domain, the number should come out as the functions range. What is the functions range?

Let's look at x^2. You can put any number in there. Thus, the Domain is the set of Real Numbers, which is any number. Can you get a negative from x^2? No, you cannot. Thus, the Range would be [0,infinity).

For the question:

lim(x-3)/(x-2)
x->2+

Is it defined when x=2? No, it is not. Is it defined for every other number? Yes, it is. Therefore, the Domain is (-infinity,2)U(2,infinity). In other words, the function is continuous everywhere, but not at x=2.

For that question, I believe you can only solve it numerically. In other words, with a calculator by taking values very close to 2. i.e. 2.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 :) This way you know for sure it is right!

I'm kidding there. Something like 2.0001 should be sufficient.

8. Oct 18, 2004

### UrbanXrisis

Is JonF saying that whenever the denominator approaches 0, the limit would be infinity?

9. Oct 18, 2004

### JasonRox

Yes, he is.

lim1/x=infinity
x->0

Try 1/0.001 and then 1/0.000001.

The numbers get .... bigger ... and bigger.

10. Oct 18, 2004

### JasonRox

Note:

Infinity is not a number.

11. Oct 19, 2004

### HallsofIvy

Staff Emeritus
In general, if the limit of the numerator is not 0 but the limit of the denominator is 0, then the limit is "infinity". I put that in quotes because, as JasonRox said, infinity is not a number. That's really just a way of saying that that the limit does not exist (in a particular way).

12. Oct 19, 2004

### JasonRox

For that reason alone, here is the guideline to whether or not a limit exists.

The limit must be a finite number.

13. Oct 19, 2004

### Shay10825

What if the limit of the numerator is 0 then what?

14. Oct 19, 2004

### JasonRox

If the denominator is not zero, then it equals zero.

As x approaches 0 for x/5, it gets closer and closer to 0. Therefore, the limit is zero. It does exist since 0 is a finite number.

If it looks something like this x/x and x approaches 0, then it equals to 1.

NOTE: It is very important to note that it is not 0/0 equals 1 because x never equals to 0, but only approaches zero. Therefore, 0.0001/0.0001 = 1.

15. Oct 19, 2004

### Shay10825

Well when does a limit not exist (when is it not inf and the answer is does not exist)? Would every limit be a number or inf? Also how do you know if a limit is pos inf or neg inf?

16. Oct 19, 2004

### JasonRox

What does "inf" mean?

A limit exists when it is defined as a finite number.

Also, for the limit as x approaches a for the function f(x). The left-hand limit must equal the right-hand limit.

If you get 5 coming from the left of a, and 8 coming from the right, you can not conclude that the limit at a is b (b is constant - finite number). Why? Because 5 and 8 do not equal. You can't choose one or the other. In this case, you say it does not exist and explain why. You need to show that the left-hand and right-hand limit do not equal.

Look at the graph below.

--------o............
...........*---------

It's kind of lame, but it will give you an idea. Ignore the dots. Know when approaching from the left the value is b, but approaching from the right it equals b-1. They do not equal.

Look at this graph.

-------o--------

Even though it is not defined at a (or o), we know the limit is b.

Can you tell me why that is?

17. Oct 19, 2004

### Shay10825

inf means infinity

18. Oct 19, 2004

### Shay10825

Because it approaches the same number from each side?

19. Oct 19, 2004

### Shay10825

Why is the answer the limit does not exist? How would I know this without a calculator and without graphing them?

lim 1+ (1/x)
x->0-

and

lim 2/sinx
x->0+

20. Oct 19, 2004

### JasonRox

For the first one, it should be negative infinity.

I explained earlier why that is.

In other words...

lim 1/x
x->0

...does not exist. First of all, the left and right don't equal. One is positive infinity and the other is negative infinity. Second of all, infinity is not finite so it doesn't exist.

Note: Just because it doesn't exist does not mean we must ignore that it equals infinity. The fact that it equals infinity tells us how the graph may look like and how the function behaves when x approaches 0.

For the second limit, take a wild guess.

Hint: It is VERY similiar to the first.