Continuous Functions 1

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- anonymous

Continuous Functions 1

- jamiebookeater

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- anonymous

Check if the function is continuous:
\[f(x) =\] {
\[{x^2 - 1, x < 1}\]
\[4-x, x \ge 1\]
}

- anonymous

Okay - you'll see that the part of f that you use to evaluate the function will depend on whether x is less than 1, or greater than or equal to 1. To show that the function is continuous, you're being asked to show that the value you get for f(x) and you approach 1 from either side, is the same. I'll show you.

- anonymous

brb

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- anonymous

ok, tks!

- anonymous

So this is about taking things called, "left" and "right" limits. When we approach a function from the left, it's a left-hand limit. When it's from the right, it's a right-hand limit.
When x<1, you to get to 1, you need to approach f(x) from the left. So we write\[\lim_{x^- \rightarrow 1}f(x)=\lim_{x^- \rightarrow 1}(x^2-1)=1-1=0\]

- anonymous

This is because f(x)=x^2-1 is only defined for x<1.

- anonymous

We couldn't use f(x) = 4-x since this is only true for x>=1.

- anonymous

*Now* we look at the right-hand limit.

- anonymous

I begin to understand.

- anonymous

\[\lim_{x^+ \rightarrow 1}f(x)=\lim_{x^+ \rightarrow 1}(4-x)=4-1=3\]

- anonymous

Now, the function can only be continuous if it has the same value when x approaches 1 from either side (otherwise, if you look at a plot, the graph will be cut in two).
Since your function does not reach the same value as we approach from the left and the right, it cannot be continuous. If you were to plot f(x), you could see it visually.

- anonymous

UNDERSTAND!
0 <> 3.
I have another problem of that kind, I'll try to do it. Thank you!

- anonymous

Excellent, I'm glad! Become a fan :p

- anonymous

I am already!
:)

- anonymous

:D

- anonymous

Ah... What kind is this discontinuity? Removable, infinite, ...?

- anonymous

First kind, irremovable discontinuity.

- anonymous

Otherwise known as a 'jump' discontinuity.

- anonymous

Thanks again!

- anonymous

np

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