Ask
your own question, for FREE!
Mathematics
14 Online
.
Still Need Help?
Join the QuestionCove community and study together with friends!
yea im alright
i heard of them
what do you got for a question?
nope
i mean nope as in no limit
Still Need Help?
Join the QuestionCove community and study together with friends!
yes
the function is really \[ f(x) = \left\{ \begin{array}{lr} -1 & : x <\frac{1}{2}\\ 1& : x >\frac{1}{2} \end{array} \right.\]
if \(x<\frac{1}{2}\) then \(|2x-1|=1-2x\) and \[\frac{|2x-1|}{2x-1}=\frac{1-2x}{2x-1}=-1\]
similarly if \(x>\frac{1}{2}\) then \(|2x-1|=2x-1\) and \(\frac{2x-1}{2x-1}=1\)
so it really looks like this |dw:1411351139323:dw|
Still Need Help?
Join the QuestionCove community and study together with friends!
and the limit from the left is \(-1\) the limit from the right is \(1\) and since \(-1\neq 1\) there is no limit
When working with limits, it helps to look at the graph of the functions given.
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
Arriyanalol:
@tinydinoUwU stop trying to find a argument u blad lil boy
TinydinoUwU:
**(Verse 1)** Yo, trapped in a box, Iu2019m feelin' so confined, Lifeu2019s a game of chess, but Iu2019m stuck in rewind, Every dayu2019s a struggle, man, I
Arriyanalol:
hey umm so i need help with my lanauage art ixl anybody wanna help big mama
Nina001:
ho where do i go to buy Subscirption for a moving pfp because on my screen im on
1 day ago
5 Replies
4 Medals
9 hours ago
13 Replies
5 Medals
4 days ago
2 Replies
2 Medals
5 days ago
4 Replies
2 Medals