Question

limits:O

Answer 1

\[\lim_{x \rightarrow 5} \frac{ x-5 }{ \left| x-5 \right| }\]

Answer 2

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?

Answer 3

Erm, yes.

Answer 4

Okay. Then coming in from the right at 5, what is your limit equal?\[\lim_{x \rightarrow 5^+} \frac{ x-5 }{ \left| x-5 \right| }=?\]

Answer 5

0/0

Answer 6

oh wait. \[infinity\]

Answer 7

would it be DNE?

Answer 8

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.

Answer 9

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?

Answer 10

okay. Wait each side would be infinity right?

Answer 11

Try plugging in values I showed you. Also, a note for the prior statement I made: f has to be continuous.

Answer 12

at 6 it would be -1/2

Answer 13

right?:3

Answer 14

I read that it has to be either DNE, infinity, or negative infinity

Answer 15

@vf321 sorry, i'm confused :(:#

Answer 16

Lets try this again.\[\frac{6-5}{|6-5|}=1\] how on earth did u get 1/2?

Answer 17

I was accidentally looking at a different problem:3

Answer 18

okay. now try 5.0001

Answer 19

But okay, then 5.0001 = 1 too.

Answer 20

okay. try 5.000000000000000001

Answer 21

Anything >5 will be 1

Answer 22

good. In fact, you can say\[\lim_{x \rightarrow 5^+} \frac{ x-5 }{ \left| x-5 \right| }=1\]

Answer 23

Now try\[\lim_{x\rightarrow 5^-}f(x)=?\]From the left, on your own.

Answer 24

o.o okay that makes sense. Let me try...

Answer 25

-1

Answer 26

So it's DNE, because they are different.

Answer 27

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.

Answer 28

Do I write that?:O

Answer 29

no but its for ur understanding.

Answer 30

Oh okay, thank you:)

Answer 31

remeber to close the question.