Re: Another limit problem

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**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 x*_{0} if f(x) gets closer to L as x approaches x_{0}.

Explain why the function in your example does not have the given value of L as a

limit as x -> x_{0}.

Consider $\displaystyle f(x)=(x-2)^2+1$ and $\displaystyle x_0=2$.

Now the closer $\displaystyle x$ is to $\displaystyle 2$, the closer $\displaystyle f(x)$ is to $\displaystyle 0$.

But $\displaystyle {\lim _{x \to 2}}f(x) = 1 \ne 0$

Re: Another limit problem

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**Plato** Consider $\displaystyle f(x)=(x-2)^2+1$ and $\displaystyle x_0=2$.

Now the closer $\displaystyle x$ is to $\displaystyle 2$, the closer $\displaystyle f(x)$ is to $\displaystyle 0$.

But $\displaystyle {\lim _{x \to 2}}f(x) = 1 \ne 0$

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.

Re: Another limit problem

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.

Re: Another limit problem

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.

Re: Another limit problem

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**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 $\displaystyle f$ is $\displaystyle f(2)=1$ so the closer $\displaystyle x$ is to $\displaystyle 2$ the closer $\displaystyle f(x)$ is to $\displaystyle 0$.

Re: Another limit problem

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

(Bow)

-Dan