Question

write a function so that as the limit of x approaches 0 of f(x) it doesnt exist, but as the limit of x approaches 0, of the absolute value of f(x) exists

Answer 1

make a piecewise function might work

Answer 2

write a function so that \[\lim_{x \rightarrow 0}f(x)\] doesnt exist, but
\[\lim_{x \rightarrow 0} \left| f(x) \right|\] exists

Answer 3

thats wat i thought, but i cant think of one as x approaches 0, maybe for 2 or 3, but not 0 :(

Answer 4

you want a simple answer, or an idea?

Answer 5

an answer, and explanation maybe? ive been stuck on this problem for 3 days :(

Answer 6

ok first of all, all functions you know are continuous on their domain
all trig functions, exponential functions, rational functions logarithmic function etc etc

Answer 7

so if you want to come up with an example of a function where the limit at 0 or any other number does not exist, your best bet is to come up with a piecewise function, defined one way for \(x<0\) and another way for \(x>0\)

Answer 8

you know what i mean when i say a "piecewise" function?

Answer 9

"no" is a fine answer, i am just asking
i can explain if you like

Answer 10

yes i do, its usually used with absolute value functions, or filling in a removable discontinuity

Answer 11

here is an example
\[f(x) = \left\{\begin{array}{rcc}
x + 4 & \text{if} & x <0 \\
- x - 4& \text{if} & x\geq 0
\end{array}
\right.
\]

Answer 12

wouldnt the top one be > rather than <

Answer 13

as a matter of fact, that might be an example you could use as an answer
i was going to give a much simpler one
so simple you are going to be annoyed that you spent three days or even three minutes on it

Answer 14

no they can't both be greater than

Answer 15

no, i mean the bottom one would be < and equal to, while the top one is >

Answer 16

how about this simple function
\[f(x) = \left\{\begin{array}{rcc}
-1& \text{if} & x <0 \\
1& \text{if} & x\geq 0
\end{array}
\right.
\]

Answer 17

it is my function i can write it any way i choose

Answer 18

........... but then the function wouldnt be continuous with that function

Answer 19

i mean, limit wouldnt exist

Answer 20

write a function so that as the limit of x approaches 0 of f(x) it doesnt exist,

Answer 21

if that is the case, it cannot be continuous

Answer 22

because the left hand limit wouldn't equal the right hand limit

Answer 23

\[f(x) = \left\{\begin{array}{rcc}
-1& \text{if} & x <0 \\
1& \text{if} & x\geq 0
\end{array}
\right.\] the left handed limit is \(-1\) and the right handed limit is \(1\)

Answer 24

and one doesn't equal -1, also, there needs to be a non absolute value function where the limit wouldnt exist

Answer 25

exactly

Answer 26

clear or no?
first it says
\(\textbf{"write a function so that as the limit of x approaches 0 of f(x) it doesnt exist"}\) i say
\[f(x) = \left\{\begin{array}{rcc}
-1& \text{if} & x <0 \\
1& \text{if} & x\geq 0
\end{array}
\right.\] fits that bill

Answer 27

o wait, i get it now

Answer 28

by being absolute value, both sides would be 1 rather than -1 and 1 o.o

Answer 29

then it says
but as the limit of x approaches 0, of the absolute value of f(x) exists
i say \[f(x) = \left\{\begin{array}{rcc}
-1& \text{if} & x <0 \\
1& \text{if} & x\geq 0
\end{array}
\right.\] als fits that bill because
\[|f(x)|=1\]

Answer 30

thank you, u were right, a simple function solved 3 days of effort T_T

Answer 31

make up your own example now

Answer 32

ok

Answer 33

\[f(x) = \left\{\begin{array}{rcc}
x + 4 & \text{if} & x <0 \\
- x - 4& \text{if} & x\geq 0
\end{array}
\right.\] might also work, you can check it