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# Explain why the function is discontinuous at the given number $a$. Sketch the graph of the function.$f(x) = \left\{ \begin{array}{ll} \cos x & \mbox{if$ x < 0 $}\\ 0 & \mbox{if$ x = 0 $} \hspace{42mm} a = 0\\ 1 - x^2 & \mbox{if$ x > 0 $} \end{array} \right.$

## $\lim _{x \rightarrow 0} f(x)=1,$ but $f(0)=0 \neq 1,$ so $f$ is discontinuous at 0

Limits

Derivatives

### Discussion

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##### Top Calculus 1 / AB Educators  ##### Samuel H.

University of Nottingham ##### Michael J.

Idaho State University Lectures

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### Video Transcript

Okay let's explain why the function is discontinuous that they give a number a let's sketch a graph. Okay let's sketch this graph. Right? So if X is less than zero then the F. Of X. Is equal to co sign next. So let's grab that. Two pi negative three pi halves negative pi negative pi house and zero. So on the left hand side when X is less than zero this is just co sign. Where does coastline start? Carson always starts at one drops down to negative, Dropls down to zero And that negative one and then one and then zero. Sorry listen this up then back up again. Nice little lotus like flower shape. You should know what the graph of coastline looks like. Okay now I missed a little tiny stuff that's pretty important here which is Co sign is less than zero when it's less than zero not less than or equal to. Which means there has to be an open circle here. This isn't even more important because the next piece of the piecewise is that f of X equals zero. If it's if X is equal to zero so at X equals zero we're going to have a point right here at zero. Okay if X is greater than zero and we're going to have the function one minus escort. So let's plug zero in for their if we plug zero in for one minus X squared We're going to get 1 0 which is one. So let's not do that. Let's graph the function a little further on. Um mm. Black. Okay I don't really want to do this in terms of pie anymore because that's kind of a pain, right? And with this the scale of the graph doesn't matter as much. But let's let's try pie halves is about 1.7. So I take that, that's about one two, three for that should be enough. Okay let's do the right most side now one minus X squared. If I plug in one I'm going to get zero. If I plug in zero. This is important to I'm going to get one. But this is also an open circle because it says X is greater than zero, not X is greater than or equal to zero. The only place where I can place the only spot at zero where I can place a closed circle like a dot of value is zero because that is what the piece by says zero if X equals zero. The other functions don't let me do that. So now let's plug if we plug in the one we get zero. If we plug in the two we get negative three because two squared Is 4, 1 -4 is -3. It'll just keep going down three Squared is nine. Yeah it's going to be way down there. So there will be something like we yep it's like a downward parabola. So it should actually look more like that. More of a bell shape than like that. Three should be somewhere there. Okay now let's talk about why it's discontinuous at equal zero. Let's take the limits. The limit As X approaches zero from the left of F of X is going to refer to a certain function in the piecewise zero from the left means less than zero. So this is cosine X As it approaches zero from the left. If we plug in zero further we get one. But if we plug in, yeah, zero from the right, we're going to get F of X greater than zero X plus of one minus X squared. We also get one. So in this situation this is a removable dis continuity because the limit from the left which is the co sign peace is equal to the limit from the right at one. Therefore the limit As X approaches zero of F of X exists however, And this is important here. The limit as X approaches zero of F of X is not equal to F of X. So the limit as X approaches zero from Alphabet's one. But the value as we see in the piecewise is zero. So it's discontinuous because of a removable this continuity. That's what we call this situation. Removable discontinuity when the limits from the left and the right exist, but they don't match the value of the function itself. That's a removable discontinuity. Yeah. Yeah

AK
The University of Alabama

#### Topics

Limits

Derivatives

##### Top Calculus 1 / AB Educators  ##### Samuel H.

University of Nottingham ##### Michael J.

Idaho State University Lectures

Join Bootcamp