Limit does not exist but function exist

In summary, the conversation discusses a case in which the limit of a piecewise defined function does not exist, but the value of the function at that point does exist. The conversation also includes a discussion of the four different ways a function can be represented and suggests that considering other representations may be helpful in understanding the concept.
  • #1
funlord
15
1

Homework Statement


there are four cases on limits given to us, and one of them I didnt really understand.
This case was: Limit f(x) as x approaches a does not exist but f(a) exist.

Homework Equations


upload_2015-6-20_8-32-29.png

The Attempt at a Solution


My answer here that the limit in this piecewise defined function exist,
As x approaches -5 from the left, it uses the function x+3;
And as x approaches -5 from the right, it uses the function √(25-x^2)please give me a great explanation with this case

Thank you
 
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  • #2
funlord said:

Homework Statement


there are four cases on limits given to us, and one of them I didnt really understand.
This case was: Limit f(x) as x approaches a does not exist but f(a) exist.

Homework Equations


View attachment 85012

The Attempt at a Solution


My answer here that the limit in this piecewise defined function exist,
As x approaches -5 from the left, it uses the function x+3;
And as x approaches -5 from the right, it uses the function √(25-x^2)please give me a great explanation with this case

Thank you

Do you really think that 0 = -2? That would need to be true in order that f(x) have a limit as x → -5.
 
  • #3
funlord said:
As x approaches -5 from the left, it uses the function x+3;
This suggests that the limit should be -2.

funlord said:
And as x approaches -5 from the right, it uses the function √(25-x^2)
This suggests that the limit should be 0.

A limit of f at 5 is a number L such that for all ##\varepsilon>0## there's a ##\delta>0## such that the following implication holds for all real numbers ##x##.
$$0<|x-5|<\delta\ \Rightarrow\ |f(x)-L|<\varepsilon.$$ The implication is saying that f maps the interval ##(5-\delta,5+\delta)## into the interval ##(L+\varepsilon,L-\varepsilon)##.

Now let's see if there a ##\delta>0## with the property above, in the special case when ##\varepsilon=1##. No matter how small we choose ##\delta##, there's an x in the interval ##(5-\delta,5+\delta)## such that f(x) is not in the interval (-2-1,-2+1)=(-3,-1). (Choose x slightly greater than 5, so that f(x) is very close to 0). This implies that -2 can't be a limit. And there's also a y in the interval ##(5-\delta,5+\delta)## such that f(y) is not in the interval (0-1,0+1)=(-1,1). (Choose y slightly less than 5, so that f(y) is very slose to -2). This implies that 0 can't be a limit.
 
Last edited:
  • #4
funlord said:

Homework Statement


there are four cases on limits given to us, and one of them I didnt really understand.
This case was: Limit f(x) as x approaches a does not exist but f(a) exist.

Homework Equations


View attachment 85012

The Attempt at a Solution


My answer here that the limit in this piecewise defined function exist,
As x approaches -5 from the left, it uses the function x+3;
Yes, and so as x gets closer and closer to -5 (from the left), f(x) gets closer and closer to -5+ 3= -2.

And as x approaches -5 from the right, it uses the function √(25-x^2)
Yes, and so as x gets closer and closer to -5 (from the right), f(x) gets closer and closer to √(25-(-5)^2)= 0.
But we can't say that f(x) gets closer and closer to anyone number as x gets closer and closer to -5.

please give me a great explanation with this case

Thank you
 
  • #5
thank you for the great explanation
 
  • #6
There is a typo on that page the image is taken from:

upload_2015-6-23_12-0-36.png


It seems this should have said:

##\displaystyle\ \lim_{x\to -5}\, f(x) \ \text{ does not exist. [why?]} \ ##​

As it's stated in your image, the limit does exist. It's simply the same as the constant ƒ(-5) , which is zero .
 
  • #7
funlord said:

Homework Statement


there are four cases on limits given to us, and one of them I didnt really understand.
This case was: Limit f(x) as x approaches a does not exist but f(a) exist.

Does the problem say "there exists an a such that..." or "for all a.."

Also, are you required to give an equation as the representation of a function? A function can be represented in 4 different ways: in words, as an equation, in a table, and as a graph. Perhaps reasoning with one of these other representations may be helpful.
 

FAQ: Limit does not exist but function exist

1. What does it mean when the limit does not exist but the function exists?

When the limit does not exist but the function exists, it means that the values of the function do not approach a specific number as the independent variable approaches a certain value. This could happen when there is a jump or a discontinuity in the graph of the function.

2. Can a function exist without a limit?

Yes, a function can exist without a limit. This means that the function has a well-defined value at a specific point, but the values of the function do not approach a specific number as the independent variable approaches that point.

3. How do you determine if a limit does not exist but the function exists?

To determine if a limit does not exist but the function exists, you can graph the function and observe if there is a jump or a discontinuity at the point in question. You can also use algebraic techniques, such as taking the limit from the left and right sides of the point, to see if they approach different values.

4. What is the difference between a discontinuous function and a function with a limit that does not exist?

A discontinuous function is a function that has a break or a gap in its graph, while a function with a limit that does not exist does not have a well-defined value at a certain point. In other words, a discontinuous function has a point of discontinuity, while a function with a limit that does not exist has a point of non-existence.

5. Can a function have a limit that does not exist at multiple points?

Yes, a function can have a limit that does not exist at multiple points. This could happen when there are multiple jumps or discontinuities in the graph of the function. In this case, the function may still have a well-defined value at each point, but the limits approaching those points do not exist.

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