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# If $H$ is the Heaviside function defined in Example 2.2.6, prove, using Definition 2, that $\displaystyle \lim_{t \to 0} H(t)$ does not exist. [$Hint:$ Use an indirect proof as follows. Suppose that the limit is $L$. Take $\varepsilon = \frac{1}{2}$ in the definition of a limit and try to arrive at a contradiction.]

## $\lim _{t \rightarrow 0} H(t)$ doesn't exist

Limits

Derivatives

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##### Top Calculus 1 / AB Educators    ##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

So in this problem were given the hughesside function, H. O. T. Is zero. If T. Is less than zero and one If T. is greater than or equal to one. And were asked two. Use definition to and prove that approved That the limit as T approaches zero. Ah H. Of T does not exist. Okay so we're gonna use what what we call an indirect proof. Right? So we're going to I suppose that the limit of H. Of T. As T goes to zero exists. Yeah. And so we're going to write the limit as T. Goes to zero of H. Of T. Is L. So then by definition a definition For all eps long greater than zero. There exists A Delta Greater than zero. Search the hat. Such that what happens? Well such that H of t minus L. Is less than epsilon. Whenever t zero is less than Delta. Since delta script zero. This is also greater than zero. Okay now let's suppose ε is one half. So then we have three cases. First one Cheese less than zero. Right? Remember when this happens by definition Hft is zero. So we got a church of t minus L less than a half. Whenever zero is less than t minus zero minus has less than delta. Okay well Hft is zero. So that means I have I have the absolute value of minus L. Less than a. Uh huh. Whenever well T is also zero. Right okay so I'm the value of T less than delta. Okay so what does that mean? Well that means that I end up with L less than 1/2, Don't I? Okay now let's look at the next case We got T greater than zero then H of T. Is one isn't it? When this happens. So by the same logic, H A T minus L. It's less than one half. Whenever zero is less than T zero is less than delta. Okay So what does that mean? Well Hft is one. So that means I have I am survey of one -L is less than a half whenever Well T is greater than zero. So this ups the value of T less than delta. Okay so This means what well this means I have 1 -6 less than a half. Whenever absolute value of T. Is less than delta. So this means that L now is greater than I have because I subtract one to the other side that gives me minus a half on the right. And then I multiplied by minus to get L. I mean I have to switch to inequality around so you know is less than observe a iot is less than delta. Okay. And our third case we'll let T equals zero. Now then by definition Hft is one right by definition of our function. And so the limit His T goes to zero of H. of T. If this is L. Then uh huh. Then the limit As T goes to zero from the left of H of T past equal the limit as T. Goes to zero from the right of H. O. T. Which is the two limits that we just did. And we saw from above that this this first limit was well we know that When T is less than zero which is which is here right, this limit is going to be zero because HFT is zero over here. So we have a limit As T goes to zero from the left Um H. of T. is zero. And the limit as T goes to zero from the right of Hft a definition of HFT is one. Okay? And these are not equal, are they? So that means that right? All three cases now that the limit as T. Goes to zero of H. Of T. Does not exist because we couldn't find it when the limit went to T. Or T. We went to zero. And the limit for T less than zero. Yes, less than a half. And the limit for T greater than zero is greater than a half. So none of these limits are equal. Therefore this limit does not exist

DM
Oklahoma State University

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Limits

Derivatives

##### Top Calculus 1 / AB Educators    ##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp