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hawkfalcon
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\[\lim_{x \rightarrow 5} \frac{ x-5 }{ \left| x-5 \right| }\]
vf321
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Well, do you know that\[\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a^+}f(x)\cap\lim_{x\rightarrow a^-}f(x)\]I.e., for 1 variable, the bilateral limit exists iff both lateral limits exist and are the same?
hawkfalcon
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Erm, yes.
vf321
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Okay. Then coming in from the right at 5, what is your limit equal?\[\lim_{x \rightarrow 5^+} \frac{ x-5 }{ \left| x-5 \right| }=?\]
hawkfalcon
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0/0
hawkfalcon
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oh wait. \[infinity\]
hawkfalcon
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would it be DNE?
vf321
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The right-side limit is not DNE. When solving limits you can only say that\[\lim_{x\rightarrow a}f(x)=f(a)\] if f(a) exists.
vf321
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Think about it. We're approaching the limit from the right. When x = 6, what's the limit? When x = 5.01, what's the limit?
hawkfalcon
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okay. Wait each side would be infinity right?
vf321
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Try plugging in values I showed you. Also, a note for the prior statement I made: f has to be continuous.
hawkfalcon
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at 6 it would be -1/2
hawkfalcon
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right?:3
hawkfalcon
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I read that it has to be either DNE, infinity, or negative infinity
hawkfalcon
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@vf321 sorry, i'm confused :(:#
vf321
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Lets try this again.\[\frac{6-5}{|6-5|}=1\] how on earth did u get 1/2?
hawkfalcon
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I was accidentally looking at a different problem:3
vf321
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okay. now try 5.0001
hawkfalcon
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But okay, then 5.0001 = 1 too.
vf321
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okay. try 5.000000000000000001
hawkfalcon
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Anything >5 will be 1
vf321
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good. In fact, you can say\[\lim_{x \rightarrow 5^+} \frac{ x-5 }{ \left| x-5 \right| }=1\]
vf321
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Now try\[\lim_{x\rightarrow 5^-}f(x)=?\]From the left, on your own.
hawkfalcon
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o.o okay that makes sense. Let me try...
hawkfalcon
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-1
hawkfalcon
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So it's DNE, because they are different.
vf321
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yes! and, if you were to graph the function on a CAS (or Wolfram|Alpha), you'd see there's a jump discontinuity at 0.
hawkfalcon
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Do I write that?:O
vf321
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no but its for ur understanding.
hawkfalcon
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Oh okay, thank you:)
vf321
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remeber to close the question.