Another limit problem

**Nora314** · September 2nd 2012, 11:26 AM

Hello everyone!

So, this is the question, and I am not sure I am able to find any reason why this statement is wrong, as the exercise requires.

Exercise 2.3.53

Show by example that the following statement is wrong: The number L is the limit of
f(x) as x approaches x₀ if f(x) gets closer to L as x approaches x₀.
Explain why the function in your example does not have the given value of L as a
limit as x -> x₀.

I have just started my university calculus course and am feeling pretty lost most of the time. Some words of advice would be nice if anyone has any hehe.

Thank you everyone!

**Plato** · September 2nd 2012, 11:58 AM

Originally Posted by Nora314

Exercise 2.3.53
Show by example that the following statement is wrong: The number L is the limit of
f(x) as x approaches x₀ if f(x) gets closer to L as x approaches x₀.
Explain why the function in your example does not have the given value of L as a
limit as x -> x₀.

Consider $f(x)=(x-2)^2+1$ and $x_0=2$ .
Now the closer $x$ is to $2$ , the closer $f(x)$ is to $0$ .
But ${\lim _{x \to 2}}f(x) = 1 \ne 0$

**Prove It** · September 2nd 2012, 07:31 PM

Originally Posted by Plato

Consider $f(x)=(x-2)^2+1$ and $x_0=2$ .
Now the closer $x$ is to $2$ , the closer $f(x)$ is to $0$ .
But ${\lim _{x \to 2}}f(x) = 1 \ne 0$

Are you sure about that? If I was to approach 2, I'm sure my f(x) would approach 1.

I think the problem with the statement as given is that you need to approach L FROM BOTH SIDES as you make x approach x_0 FROM BOTH SIDES in order to have L be the limit of f(x) as x approaches x_0.

**SworD** · September 2nd 2012, 09:53 PM

The idea is that you need to not only approach it, but it is also required that you get arbitrarily close to the limit. In the example stated above, in a manner of speaking, you are "getting closer to" 0 from both sides, but you never get closer than 1 unit away. So the limit is not 0.

**Nora314** · September 2nd 2012, 11:09 PM

Thanks everyone for the reply! It is much more clear to me now.

This was a bit of a funny question, seems like a lawyer question and not math hehe.

**Plato** · September 3rd 2012, 02:12 AM

Originally Posted by Prove It

Are you sure about that? If I was to approach 2, I'm sure my f(x) would approach 1.

Yes, I am quite sure of it. The minimum value of $f$ is $f(2)=1$ so the closer $x$ is to $2$ the closer $f(x)$ is to $0$ .

**topsquark** · September 3rd 2012, 04:00 AM

Originally Posted by Plato

Yes, I am quite sure of it. The minimum value of $f$ is $f(2)=1$ so the closer $x$ is to $2$ the closer $f(x)$ is to $0$ .

-Dan

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