# Sketching Graphs using limit information

1. May 31, 2013

### Physherman

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

For each of the four cases below, sketch a graph of a function that satisfies the stated conditions. In each case, the domain of the function should be all real numbers. (professor also mentioned he wants us to write it out in piecemeal function format)

a) lim(x→2) of f(x) = 3 and f(2) = 4
b) lim(x→0) of f(x) = DNE, and |f(x)| < 2 for all x
c) lim(x→1) of f(x) exists and its value is f(1) + 2
d) lim(x→-1[from the right]) of f(x) and lim(x→-1[from the left]) = DNE, |f(x)| < 3 for all x, and f(-1) = -2

2. Relevant equations

Not Applicable

3. The attempt at a solution

I am completely at a loss, but:

a) Greatest integer function?
b) -1/x^2-3
c) no idea
d) no idea

2. Jun 1, 2013

### Simon Bridge

It reads that any function that satisfies the description will do.

You can test each of your guesses by applying the limit to see if it pans out.
It's basically testing if you know what it means to take the limit.

3. Jun 1, 2013

### Quinzio

For some of the items you may find helpful that $$sin(f(x))$$ is a bounded function.

4. Jun 1, 2013

### Simon Bridge

$\sin f(x)$ :)

I think it is simpler than that though - awaiting confirmation from OP but:
- the function does not have to be named - no equation is asked for - just a sketch.
A random squiggle (provided each x only has one y) that fits the description will do.
The function does not even have to be continuous.

So, if a condition is that |f(x)|<a, then draw a dotted line at y=a and y=-a.
Another condition may be that f(p)=q ... then one places a dot at (x,y)=(p,q).
A line through the dot that does not touch the dotted lines will satisfy the conditions - leaves the limits.
I'd draw a vertical dotted line through the limit to help decide what happens there.

So.
a. what does it mean when $\lim_{x\rightarrow a}f(x)\neq f(a)$ ?
b. what does it mean that the limit does not exist?

c & d are cunning variations on a & b.

Mind you - being able to produce an equation would slam-dunk the question.

5. Jun 1, 2013

### Physherman

Simon, you are right that it can be any function.

I think I am having a very fundamental misunderstanding (still the first week of class, and I haven't taken precalc in 12 years.

I don't understand how the limit as x approaches 2 can be 3, but f(2) = 4

6. Jun 1, 2013

### Simon Bridge

Have a look at these three functions...

[Broken]

... in each case, what is the limit a x --> a from the left and from the right?
Compare, in each case, with the value of f(a).

http://scides.ca/courses/calculus_12/course/unit2/U02L04.htm [Broken]

Last edited by a moderator: May 6, 2017
7. Jun 1, 2013

### Physherman

a) totally makes sense now, it is basically graph 3.

b) since the limit does not exist as x -> 0, 1/x^2 + 2 would satisfy - because it is absolute value of f(x), 1/x^3 + 2 would also work.

Am I on the right track here?

Thanks again for the guidance.

8. Jun 1, 2013

### Simon Bridge

$$\lim_{x\rightarrow 0}\frac{1}{x^2} = \infty$$

Take another look at graph 2.
The limit exists, even though there is no value for f(a) because the left hand and right hand limits are the same.

On graph 3, what would happen to the limits if the red curve coming in from the left ended up at the topmost point instead?

Did you try following the link?

Aside:
$$\frac{1}{x^2}\neq |x|$$.. and, anyway, $$\lim_{x\rightarrow 0}|x| = 0$$

9. Jun 2, 2013

### Physherman

Ok, I think this has been helpful. I see two of my errors now.

Would it be correct to say that the lim x->0 of 1/x^3 = DNE?

And I'm still unsure how the absolute value component is affecting the answer for b and d.

10. Jun 2, 2013

### Simon Bridge

Anything where the left hand limit disagrees with the right-hand limit.
A step function will do. You can google "the limit does not exist" and look at pictures for examples.
There's also more on limits in the website I linked - please read the links people give you, we do that to save typing.

For the other thing:

|f(x)| < a just means that -a < f(x) < a.

eg. |sin(x)| < a is true as long as a > 1 :)
trivially, if f(x) is the horizontal line through the origin, then we can confidently use it as an example of |f(x)| < 2 .

In your sketch you draw a dotted, horizontal, line at y=a and y=-a and make sure that your wiggly line representing the sketched function stays between them.