Question

The signum (or sign) function, denoted by sgn, is defined by
sgnx={−1 if x<0; 0 if x=0;1if x>0}

Answer 1

Find each of the following limits or expalin y it does not exist

a) lim x appraches 0 from teh right (sgn x)
b) lim x approaces 0 from teh left (sgn x)
c) lim x appraches 0 (sgn x)
d) lim x approaches 0 (absolute value of sgn x)

Answer 2

a) \[\lim_{x \rightarrow 0+} sgn x \]
b) \[\lim_{x \rightarrow 0-} sgn x\]
c)

Answer 3

\[c) \lim_{x \rightarrow 0} sgn x\]
\[d) \lim_{x \rightarrow 0} \left| sgn x \right|\]

Answer 4

a) 1
b) -1
c) Does not exist because 1 does not equal -1
d) 1

Answer 5

how did u get those answers

Answer 6

Use \[ \text{sgn}(x) = \frac{x}{|x|} \] for x not 0.

Answer 7

and then wat do i do? srry im hopelss w/ limits =p

Answer 8

Just think about what happens. Let's consider a).
Limit from the right, means you approach 0 by values >0. So if you plug in ANY value x>0 in sgn(x), then you get 1.

Answer 9

So the limit must be one.

Answer 10

For x>0: \[ \frac{x}{|x|} = 1 \]

Answer 11

On the other hand, in b) you approach from the left, so you approach 0 with values <0. We have for x<0: \[ \frac{x}{|x|} = -1 \]

Answer 12

Now consider c). By the definition of the limit, for a limit to exist it needs to exist from both sides AND the respective limits need to match up! But 1 is unequal to -1.

Answer 13

oh i see RHL does not equal LHL

Answer 14

exactly

Answer 15

No consider d). Here we have \[ |\text{sgn}(x)| = \frac{|x|}{|x|} = 1 \] for all x not 0.

Answer 16

So the limit is 1 no matter from what side you approach 0.

Answer 17

can u help explain another limit problem

Answer 18

Just post it. Hit good answer first to let ppl know this is solved.

Answer 19

ok...ill post it now...its simialr to this