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Year 12s: what will be the first way you research your future options? >>

Confusing limits...

RamocitoMorales

f(x)=x^2

when

-1\le x<0

1

when

x=0

x^2

when

0<x<1

0

when

1\le x \le 2

So

Unparseable latex formula:

\lim_{x\to\0}f(x)=0

True or False?

Apparently the answer's true. But I can't understand why. Surely

f(x)=1

when

x=0

?

:hmmmm:

Clarity Incognito

Original post by RamocitoMorales

f(x)=x^2

when

-1\le x<0

1

when

x=0

x^2

when

0<x<1

0

when

1\le x \le 2

What conditions must a limit satisfy to exist?

RamocitoMorales

OP

Original post by Clarity Incognito

What conditions must a limit satisfy to exist?

:dontknow:

But I'm guessing that it can't be

1

since it is 'static' at that point, i.e

x=0

, as opposed to being an interval...I think?

Clarity Incognito

Original post by RamocitoMorales

:dontknow:

But I'm guessing that it can't be

1

since it is 'static' at that point, i.e

x=0

, as opposed to being an interval...I think?

It could have been undefined at x=0 but what's important about limits is that we're looking at what happens to f(x) as x tends to 0 from the positive side and the negative side. What happens exactly at the point x=0 is not so important. You can see the left side limit (coming from the negative side of 0) that it approaches 0 and similarly for the right side limit. For a limit to exist, it must have a left side limit and a right side limit and both must approach the same limit.

My question was ambiguous because it seemed to imply what exactly it means for something to be convergent to a limit but I assume that as you just want to compute these limits that you haven't got to that stage just yet.

Original post by RamocitoMorales

:dontknow:

But I'm guessing that it can't be

1

since it is 'static' at that point, i.e

x=0

, as opposed to being an interval...I think?

It seems that you're trying to justify the continuity of the function rather than the existence of the limit.

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