Question

Please explain lim X ->2 (1/(x-2))- (1/|x-2|) = -infinity.
I have worked out part of limit?

Answer 1

Can you re-write it using LaTex? That's what the option is there for :)
it makes it better to understand and avoid confusion when helping to answer a question.

Answer 2

\[\lim_{x \rightarrow 2} (\frac{ 1 }{ x-2} - \frac{ 1 }{ \left| x-2 \right| })\]
broke it nto tow parts to evalualte at 2
\[\lim_{x \rightarrow 2-} (\frac{ 1 }{ x-2} - \frac{ 1 }{ x-2})\]
i boke i tup ad got ininfity - infinty
\[\lim_{x \rightarrow 2+} (\frac{ 1 }{ x-2} - \frac{ 1 }{ -x+2 })\]
ininfity + inifinty

Answer 3

i used limit laws loosely, but this seems inaccurate

Answer 4

For absolute, you need to evaluate it from both sides, when x > 2 and x <2

Answer 5

and, as you know, when two limits do not approach the same from both sides, then the limit D.N.E

Answer 6

i know the limit exists, because my answer is wrong. They are both negative infinity . I just am note sure where i went wrong

Answer 7

when x>2, we get zero

when x <2 go functional

Answer 8

the limit from right is zero, the limit from the left is not 0

Answer 9

f(x) = 2/(x-2) x < 0
f(x) = 0 x > 2

Answer 10

lol, missed my piece domian ... x < 2 on top

Answer 11

I don'tunderstand any of your post amistre64

Answer 12

smh

Answer 13

i dont know how to do the fancy schmancy piecewise in latex coeds ...

Answer 14

it worst, create a table ....

Answer 15

\[\begin{cases}
& noob \text{ if } x>2 \\
& noob \text{ if } x<2
\end{cases}\]

Answer 16

values of x bigger than 2; and values of x less than 2

you will see that a 2 sided limit does not exists. if your answer is wrong then look at the question and see if its asking for a 1 sided limit instead

Answer 17

lol, now your just showing off

Answer 18

http://www.codecogs.com/latex/eqneditor.php

Answer 19

@amistre64

Answer 20

Okay, apparently I type it in wrong when i went to check my answer or something. It does say now that the limit doesn't exist. Thanks for the help, I can only medal one, so please medal the other since you were both helpful to me.

Answer 21

\[L.H.L=\lim_{x \rightarrow 2-}\left( \frac{ 1 }{ x-2 }-\frac{ 1 }{ \left| x-2 \right| } \right)\]
put x=2-h,h>0,\[h \rightarrow0 asx \rightarrow2-\]
\[L.H.L.=\lim_{h \rightarrow 0}\left( \frac{ 1 }{ 2-h-2 }-\frac{ 1 }{ \left| 2-h-2 \right| } \right)\]
\[=\lim_{h \rightarrow 0}(\frac{ -1 }{h }-\frac{ 1 }{\left| -h \right| })=\lim_{H \rightarrow0}\left( -\frac{ 1 }{h }-\frac{ 1 }{ h }\right)\]
\[=\lim_{h \rightarrow 0}\frac{ -2 }{h }\rightarrow-\infty \]
\[R.H.L.=\lim_{x \rightarrow 2+}\left( \frac{ 1 }{x-2 }-\frac{ 1 }{ \left| x-2 \right| } \right)\]
put x=2+h,h>0 ,\[h \rightarrow0as x \rightarrow2+\]
\[R.H.L.=\lim_{h \rightarrow 0}\left( \frac{ 1 }{2+h-2 }-\frac{ 1 }{\left| 2+h-2 \right|} \right)\]
\[=\lim_{h \rightarrow 0}\left( \frac{ 1 }{ h }-\frac{ 1 }{\left| h \right| } \right)=\lim_{h \rightarrow0}\left( \frac{ 1 }{h }-\frac{ 1 }{h } \right)=\lim_{h \rightarrow 0}=0\]
\[L.H.L \neq R.H.L\]
Limit does not exist.

Answer 22

Nice.

Answer 23

miss print 0 in the end.