LIMIT 2/x - 2/IxI x->0
Calculate the left limit and the right limit first.
it is 2/x - 2/IxI = 1
@liana123?? This is limits, you haven't learnt it yet
exactly @liana123??
what
\[\lim_{x \rightarrow 0} (\frac{ 2 }{ x }) -(\frac{ 2 }{ \left| x \right| })\]
Find the left limit and the right limit first
That is \(\displaystyle\lim_{x\rightarrow0^-}\frac2x-\frac2{|x|}\) and \(\displaystyle\lim_{x\rightarrow0^+}\frac2x-\frac2{|x|}\)
give him abit more hint @kc_kennylau
Recall yourself the definition of |x|: \[|x|=\left\{\begin{array}{lr}x&\mbox{if }x\ge0\\-x&\mbox{if }x<0\end{array}\right.\]
\[0/-x ^{2}\] and \[0/x ^{2}\]
Don't use LaTeX if you can't use it, ASCII can do you everything
this is all I am getting, or should I choose another way. All I kow is that the limit does not exist coz of the absolute value
@random231 I don't know how to do lol, you teach him? :P
if limit does not exists you cannot solve it can u?
I mean the common denominator is x squared, and and subtracting the two gave me a zero over x squares and a zero over minus x-squared, just as I have put in latex.
uh i am not used to this thingi am sorry i cant give u wrong answers. :((
I have the answers fro wolfaramalpha, I do not have the working.
whats the answer?
here it is attached
@random, EUREKA!
Thing is, the limit \[1/x \lim_{x \rightarrow 0}\] seems as if it does not exist. Howevr, when you compute the the limit manually in a table, you wil find that as x approaches \[0^{+}\], \[1/x\] increases without bound and become , and the oposite is true for \[0^{-}\], it decreases without bound and become \[-\]
\[- \infty\]
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