WEBVTT
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and this problem, we want to find the limit
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as XO. Purchase three of 3/4 x plus one
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one. Use that slime Delta definition of limit to
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find this limit. In other words, you want
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to show that for every up song grease zero,
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there exists a delta greater than zero such that the
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function through X over four plus one, minus 13
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over four absolute value is less than epsilon. Whenever
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X minus three, absolute value is between zero and
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Delta. No notice that are proposed to limit is
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13 over four. This is because for linear functions
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, which have the form F of X equals MX
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plus B, the limit as X approaches, si
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is always going to be, and C equals B
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, meaning we can just plug in this value were
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approaching into the function. So also note that if
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we plug that value in for X, we will
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get 13 over four. So that's kind of what
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we proposed that limit. But we need to show
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this because we're assuming we don't know this claim to
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be good with. So look at this value here
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again. So notice that when we simplify this by
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distributing this negative sign and combining the water Negative three
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thirteen 13 over four will get three X over four
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, minus nine over. For I noticed there we
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can kind of factor out 3/4. So who had
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the absolute value of 3/4 times X minus tree?
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We want to factor that out because we want to
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establish a relationship between this expression and this bound that
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we have a next my history or down that we
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have on the absolute value of X fine story.
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So notice that that term kind of appears here.
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Further, we can pull out the absolute value of
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three or four times absolutely of X minus three.
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And since three over four is already positive, the
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absolute value of that will just be three or four
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. So we have three or four times the absolute
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value of X, my history. So now it
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seems we should let Delta bi equal to Absalon,
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which ought to know by the letter a absolute divided
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by three over fourths, which is the same as
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four times up Salon over three, and we're done