Limits and Continuity question

In summary, the given information states that if a function f is continuous at x=5, and f(5) = 2 and f(4) = 3, then the limit as x approaches 2 of the function f(4x^2 - 11) must be 2. This is because the definition of continuity at a point "a" is that the limit as x approaches a of f(x) is equal to f(a). Therefore, since we are given that f is continuous at x=5 and f(5) = 2, the limit as x approaches 2 of f(4x^2 - 11) must also be 2.
  • #1
dkotschessaa
1,060
783

Homework Statement



True/False

If f is continuous at 5 and f(5) = 2 and f(4) = 3, then lim x-> 2 f(4x^2 - 11) = 2

Homework Equations



lim x-> 2 f(4x^2 - 11) = 2


The Attempt at a Solution



This turns out to be true, despite the fact that the limit evaluated without respect to the above information is 5. (4(2^2) - 11) = 4*4 - 11 =5

Clearly the above information changes something about the limit, or perhaps transforms the graph, but I'm not sure how.

Thanks!

-Dave KA
 
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  • #2
dkotschessaa said:

Homework Statement



True/False

If f is continuous at 5 and f(5) = 2 and f(4) = 3, then lim x-> 2 f(4x^2 - 11) = 2

Homework Equations



lim x-> 2 f(4x^2 - 11) = 2

The Attempt at a Solution



This turns out to be true, despite the fact that the limit evaluated without respect to the above information is 5. (4(2^2) - 11) = 4*4 - 11 =5

Clearly the above information changes something about the limit, or perhaps transforms the graph, but I'm not sure how.

Thanks!

-Dave KA

As with your examples from the previous thread, the function 1/x is undefined at x=0, but if you're to define the function such that f(0)=5, then it may be defined, but it is still discontinuous thus the limit at x approaches 0 is undefined.

Remember the rule that for a limit to be defined at a point, then the right hand and left hand side limits must both be equal. It doesn't need to be defined at that point though!

For example, [tex]\lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}\frac{(x-1)(x+1)}{x-1}=\lim_{x\to1}(x+1)=2[/tex]
despite the fact that the function is undefined at x=1 since that would give us the indeterminate form 0/0.
By the way, this graph is the same as y=x+1 except for the fact that it has a hole at (1,2).
 
  • #3
Sorry Mentallic, but this is an altogether different problem and I'm not sure how you're relating it to the previous. Probably would help if I was on board with entering equations in Latex eqations here. (working on it)

-Dave KA
 
  • #4
Let me rephrase then. Why MUST the lim x-> 2 be two just because of the additional information? I'm not sure how that changes the limit.
 
  • #5
dkotschessaa said:

Homework Statement



True/False

If f is continuous at 5 and f(5) = 2 and f(4) = 3, then lim x-> 2 f(4x^2 - 11) = 2

Homework Equations



lim x-> 2 f(4x^2 - 11) = 2

The Attempt at a Solution



This turns out to be true, despite the fact that the limit evaluated without respect to the above information is 5. (4(2^2) - 11) = 4*4 - 11 =5
The limit of what function? limit, as x goes to 2, of 4x^2- 11 is 5. But that is NOT the function you are taking the limit of. Because f is continuous at 5, lim_{x->5} f(x^2- 11= f(lim_(x->5) x^2- 11)= f(5). It is not "despite" but "because" lim x^2- 11= 5 that the lim f(x^2- 11)= 2.

Clearly the above information changes something about the limit, or perhaps transforms the graph, but I'm not sure how.

Thanks!

-Dave KA
There are 3 different functions here: f(x), x^2- 11, and f(x^2- 11). Be careful which function you are taking the limit of.
 
  • #6
dkotschessaa said:
This turns out to be true, despite the fact that the limit evaluated without respect to the above information is 5. (4(2^2) - 11) = 4*4 - 11 =5
I'm not sure what's confusing you. Specifically, what do you mean by "without respect to the above information"? Do you mean
[tex]\lim_{x \to 2} (4x^2-11) = 5[/tex]
vs.
[tex]\lim_{x \to 2} f(4x^2-11) = 2[/tex]
If so, why do you say "despite the fact"?
 
  • #7
dkotschessaa said:
Let me rephrase then. Why MUST the lim x-> 2 be two just because of the additional information? I'm not sure how that changes the limit.

Because the function is continuous at x=5 and f(5)=2.
If we only had the information that f(5)=2 then it says nothing about whether the function is continuous or not. Just because that value exists doesn't mean the limit exists.
And then if we were only given that the function is continuous at that point, then we know the limit exists at that point but we don't know the value of it.
 
  • #8
Ok, thanks for doing your best to explain. I'm almost there. "It's not you, it's me." :)

grokking...
 
  • #9
Ok, the definition of continuity at a number "a" is

A function is continuous at a number "a" if

lim x->a f(x) = f(a)

In this case we are told f is continuous at 5 so:

lim x-> 5 of f(x) = f(5)

and we're told f(5) = 2

So no matter what the equation is the limit has to be 2?

Sorry if I'm being dense!
 
  • #10
Yep!
 
  • #11
Ok, thanks. I need to start getting a grip on using Latex here...
 

FAQ: Limits and Continuity question

What is the definition of a limit?

The limit of a function is a mathematical concept that describes the behavior of a function as its input approaches a certain value. It is the value that the function approaches, or "approaches to", as the input value gets closer and closer to a specific value.

How do you determine if a limit exists?

A limit exists if the function approaches the same value from both the left and right sides of the input value. This means that the function is approaching the same y-value as the input value approaches the specific value from both directions.

What is the difference between a left and right limit?

A left limit, also known as a one-sided limit, is the value that the function approaches from the left side of the input value. A right limit, also known as a one-sided limit, is the value that the function approaches from the right side of the input value.

What is a removable discontinuity?

A removable discontinuity occurs when the limit of the function exists, but there is a hole or gap in the graph at that point. This means that the function is undefined at that point, but it can be made continuous by filling in the hole.

How do you determine if a function is continuous?

A function is continuous if its graph is one unbroken curve with no holes or gaps. This means that the function is defined and has a limit at every point in its domain. A function can also be continuous at a point if its limit at that point is equal to the value of the function at that point.

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