- #1

Shay10825

- 336

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

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

x->2+

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

x->2+

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- Thread starter Shay10825
- Start date

- #1

Shay10825

- 336

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

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

x->2+

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

x->2+

- #2

Sirus

- 574

- 2

- #3

Shay10825

- 336

- 0

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

Shay10825

- 336

- 0

- #5

JonF

- 621

- 1

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

Shay10825

- 336

- 0

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+

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:

- #7

JasonRox

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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

UrbanXrisis

- 1,197

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JonF said: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?

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

- #9

JasonRox

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lim1/x=infinity

x->0

Try 1/0.001 and then 1/0.000001.

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

- #10

JasonRox

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Note:

Infinity is not a number.

Infinity is not a number.

- #11

HallsofIvy

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- #12

JasonRox

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The limit must be a finite number.

- #13

Shay10825

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

- #14

JasonRox

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Shay10825 said:What if the limit of the numerator is 0 then what?

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

Think about it.

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

Shay10825

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- #16

JasonRox

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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

Shay10825

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JasonRox said:What does "inf" mean?

inf means infinity

- #18

Shay10825

- 336

- 0

JasonRox said:Can you tell me why that is?

Because it approaches the same number from each side?

- #19

Shay10825

- 336

- 0

lim 1+ (1/x)

x->0-

and

lim 2/sinx

x->0+

- #20

JasonRox

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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 similar to the first.

- #21

Shay10825

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- #22

JasonRox

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Shay10825 said:

By direct substitution, theorems, and definitions.

For x approaches 0, 1/x equals infinity.

We can't necessarily do that by hand since it isn't a finite number. It is a definition given in textbooks that it equals infinity.

- #23

Shay10825

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I understand now. Thanks for your help

- #24

JasonRox

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No problem.

Learn to love limits. :)

Learn to love limits. :)

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