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