Limit does not exist but function exist

[funlord]

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

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

 

[Ray Vickson]

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.

 

[Fredrik]

As x approaches -5 from the left, it uses the function x+3;

This suggests that the limit should be -2.

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.

 

[HallsofIvy]

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

 

[funlord]

thank you for the great explanation

 

[SammyS]

There is a typo on that page the image is taken from:

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 .

 

[thelema418]

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.