Limits and Continuity question

[dkotschessaa]

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

 

[Mentallic]

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, \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
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).

 

[dkotschessaa]

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

 

[dkotschessaa]

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.

 

[HallsofIvy]

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.

 

[vela]

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
\lim_{x \to 2} (4x^2-11) = 5
vs.
\lim_{x \to 2} f(4x^2-11) = 2
If so, why do you say "despite the fact"?

 

[Mentallic]

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.

 

[dkotschessaa]

Ok, thanks for doing your best to explain. I'm almost there. "It's not you, it's me." :)

grokking...

 

[dkotschessaa]

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!

 

[Mentallic]

Yep!

 

[dkotschessaa]

Ok, thanks. I need to start getting a grip on using Latex here...