Finding A Limit Graphically...1

[nycmathguy]

Exercises 1.2
21 - 24

21)

As I tend to 3 from the left and right on the x-axis, the height of the graph reaches 1. The limit is 1.

22)

As I tend to 0 from the left and right on the x-axis, the height of the graph reaches 1. The limit is 1.

23)

As I tend to 2 from the left and right on the x-axis, the height of the graph reaches 2. The limit is 2.

I am not too sure about 24.

You say?

 

[MathLover1]

21, 22, and 23 are correct

24. there is a hole at (1,4) and point at (1,2)

as x->1 then x^2+3 limit approaches 4
as x->1 then point (1,2) has limit 2

 

[nycmathguy]

21, 22, and 23 are correct

24. there is a hole at (1,4) and point at (1,2)

as x->1 then x^2+3 limit approaches 4
as x->1 then point (1,2) has limit 2

At the point (1, 4), there is a hole but y = 4.
At the point (1,2), there is no hole but y = 2.
The limit for the top part of the piecewise function
is 4 and the bottom part limit is 2. Since the limits are different as x tends to 1, the limit DNE.

Yes?

 

[HallsofIvy]

I have no idea what "as x-> 1 then point (1, 2) has limit 2" could mean!

A point has no limit- at (1, 2), x is fixed at 1, it is not tending to 1.

The answer to problem 24 is that as x goes to 1, f(x) goes to 4. The fact that f(1)= 2 is irrelevant.

 

[nycmathguy]

I have no idea what "as x-> 1 then point (1, 2) has limit 2" could mean!

A point has no limit- at (1, 2), x is fixed at 1, it is not tending to 1.

The answer to problem 24 is that as x goes to 1, f(x) goes to 4. The fact that f(1)= 2 is irrelevant.

I am obviously not ready for Calculus 1 questions.

 

[HallsofIvy]

In a "Pre-Calculus" class you would learn what "limit" means!

The definition is
"\lim_{x\to a} f(x)= L if and only if, given any \epsilon> 0 there exist \delta> 0 such that if 0< |x- a|< \delta then |f(x)- L|< \epsilon.

Basically that means that we can make f(x) as close to L as we want (less than any \epsilon) by taking x close enough to a (less than some \delta). But notice the "0< |x- a|". That's easily overlooked but very important! We take x "close enough to a" but not equal to a! For example suppose f(x) is defined to be 1 for all x except 0 and f(0)= 2. What is \lim_{x\to 0} f(x)? Because of that "0< |x- a|" we completely ignore "f(0)= 2". Everywhere around x= 0, but not at x= 0, f(x)= 1 so \lim_{x\to 0} f(x)= 1.

Of course having the limit equal to the value of the function is very useful. We call such functions "continuous". The example I just gave is continuous everywhere except at x= 0. It is a discontinuous (not continuous) function because it is discontinuous at x= 0.

 

[nycmathguy]

In a "Pre-Calculus" class you would learn what "limit" means!

The definition is
"\lim_{x\to a} f(x)= L if and only if, given any \epsilon> 0 there exist \delta> 0 such that if 0< |x- a|< \delta then |f(x)- L|< \epsilon.

Basically that means that we can make f(x) as close to L as we want (less than any \epsilon) by taking x close enough to a (less than some \delta). But notice the "0< |x- a|". That's easily overlooked but very important! We take x "close enough to a" but not equal to a! For example suppose f(x) is defined to be 1 for all x except 0 and f(0)= 2. What is \lim_{x\to 0} f(x)? Because of that "0< |x- a|" we completely ignore "f(0)= 2". Everywhere around x= 0, but not at x= 0, f(x)= 1 so \lim_{x\to 0} f(x)= 1.

Of course having the limit equal to the value of the function is very useful. We call such functions "continuous". The example I just gave is continuous everywhere except at x= 0. It is a discontinuous (not continuous) function because it is discontinuous at x= 0.

True but I am not in a precalculus course. I took precalculus in the Spring 1993 semester and got an A minus. What you see happening here, all the threads, is a review of a course I truly enjoy.