Graphing Asymptotes - Sketching f(x)

[ribbon]

Homework Statement

Both the following is true for a particular choice of function f(x):

For any ε>0 there exists an N>0 such that x>N --> |f(x) - 2| < ε
For any ε>0 there exists an N<0 such that x<N --> |f(x)| < ε

Sketch the graph of the function that satisfies both of these conditions.
There are infinitely many correct answers, you need to only sketch the graph of one of them.
Hint: can you say anything about the asymptotes of f(x)

Homework Equations

The Attempt at a Solution

I'm not sure but can it be assumed the y-axis (y=0) and y = 2 are horizontal asymptotes? And thus any graph sketched that stays in it those boundaries is acceptable .

 

[ehild]

Homework Statement

Both the following is true for a particular choice of function f(x):

For any ε>0 there exists an N>0 such that x>N --> |f(x) - 2| < ε
For any ε>0 there exists an N<0 such that x<N --> |f(x)| < ε

Sketch the graph of the function that satisfies both of these conditions.
There are infinitely many correct answers, you need to only sketch the graph of one of them.
Hint: can you say anything about the asymptotes of f(x)

Homework Equations

The Attempt at a Solution

I'm not sure but can it be assumed the y-axis (y=0) and y = 2 are horizontal asymptotes? And thus any graph sketched that stays in it those boundaries is acceptable .

The graphs need not stay between y=0 and y=2. Try to formulate the question in terms of limits. What are the limits of f(x) at +infinity and at -infinity?

ehild

 

[ribbon]

Hmmm... I would think it would be positive infinity and negative infinity respectively as limits, no?

 

[HallsofIvy]

No. "for x> N", in other words, for x very large, $|f(x)-2|<\epsilon$ So what is f(x) close to for x very large?

 

[ribbon]

No. "for x> N", in other words, for x very large, $|f(x)-2|<\epsilon$ So what is f(x) close to for x very large?

The only thing I see from that is that f(x) is within epsilon units of 2? But what should I gather from that?

 

[HallsofIvy]

Since that is true for any positive epsilon, you should gather that f(x) is very close to 2! And getting closer to 2 as x gets larger.

 

[ribbon]

Since that is true for any positive epsilon, you should gather that f(x) is very close to 2! And getting closer to 2 as x gets larger.

Ahh okay, that makes sense lim f(x) approaches 2 as x tends to ∞. But what does the other limit tell us, that when x approaches negative infinity, f(x) approaches 0?

 

[Mark44]

Ahh okay, that makes sense lim f(x) approaches 2 as x tends to ∞. But what does the other limit tell us, that when x approaches negative infinity, f(x) approaches 0?

Yes.