Thread: Question about finding the limit of a piecewise function of a piecewise function?

What is the lim as x--->2 of g(f(x)) with reference to this diagram:
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My reasoning thus far is that there is no limit because as x approaches 2 in the first graph, f(x) approaches 0. But in the second graph, of g(x), the function is not continuous at x = 0.

Is it DNE?

Originally Posted by lm1988

What is the lim as x--->2 of g(f(x)) with reference to this diagram:
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My reasoning thus far is that there is no limit because as x approaches 2 in the first graph, f(x) approaches 0. But in the second graph, of g(x), the function is not continuous at x = 0.
Is it DNE?

does not exist. DNE?

DNE = does not exist

why does it not exist?

Originally Posted by lm1988

why does it not exist?

It may exist.
What is

That is 1, from the diagram I provided, but the limit does not exist because as x approaches 0 from the left, the limit of g(x) is 3. Is this correct?

Originally Posted by lm1988

That is 1, from the diagram I provided, but the limit does not exist because as x approaches 0 from the left, the limit of g(x) is 3. Is this correct?

But the problem is .

Originally Posted by Plato

But the problem is .

Right, the problem does not specify whether x is approaching 2 from the left or the right side. On the graph of f(x) it approaches 0 from both sides, but if we approach zero on the graph of g(x), g(x) is different depending on whether we approach from the left or right side.

Originally Posted by lm1988

On the graph of f(x) it approaches 0 from both sides, but if we approach zero on the graph of g(x), g(x) is different depending on whether we approach from the left or right side.

Is that true? Look again.

Originally Posted by Plato

Is that true? Look again.

Sorry, I meant to say that on the graph of f(x) as x approaches 2 from both sides, f(x) approaches zero. However, on the graph of g(x) you cannot approach 0 from both sides with the limit being the same. Is the correct way of interpreting the question to check the value that f(x) approaches as x approaches 2 on the graph of f(x) and then take that value in order to examine what value(s) g(x) assumes?

Originally Posted by lm1988

Is the correct way of interpreting the question to check the value that f(x) approaches as x approaches 2 on the graph of f(x) and then take that value in order to examine what value(s) g(x) assumes?

That is exactly the point.
Here is a bit of notation. means that is very very close to BUT . That is the concept of limit.
If this case gives but moreover so what is true of

Originally Posted by Plato

That is exactly the point.
Here is a bit of notation. means that is very very close to BUT . That is the concept of limit.
If this case gives but moreover so what is true of

Thanks a lot! I got the answer. I never looked at it from that perspective, but since f(x) > 0 in from both the left and right side, we must approach the value g(x) thinking that the x-axis is actually the f(x) axis, meaning that we approaching f(x) = 0 from its positive side.

Originally Posted by lm1988

Thanks a lot! I got the answer. I never looked at it from that perspective, but since f(x) > 0 in from both the left and right side, we must approach the value g(x) thinking that the x-axis is actually the f(x) axis, meaning that we approaching f(x) = 0 from its positive side.

Exactly


BUT

Hey all,

This question is free game for anyone. In regards to the diagram referenced in the first post, to which I posted a link, is the value of c that makes [f(x) + c g(x)] exist equal to the same value of c that makes f(x) + c g(x) continuous at x = 1.

I approached the problem this way:

For to exist, both the left and right limits must exist.





(The question mark is there not because this limit is unknown, but I leave it to you to figure it out.) Also, the left and right limits must be equal to each other. After finding the second limit, set the two of them equal to each other, and solve for c.

Now refer to the definition of continuity to complete the question, and proceed in a manner similar to above.