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OpenStudy (anonymous):

how do you find the limit of lim x->5^+ of Ix-2I/x-2?

OpenStudy (anonymous):

\[\lim_{x \rightarrow 5^{+}} \left| x-2 \right| / x-2\]

myininaya (myininaya):

just use direct sub.

myininaya (myininaya):

|5-2|/(5-2)=3/3=1

OpenStudy (anonymous):

sryy..i meant limx→2+|x−2|/x−2

OpenStudy (anonymous):

its not a 5..its a 2

myininaya (myininaya):

if x-2>0 (x>2) then |x-2|=x-2 so |x-2|/(x-2)=(x-2)/(x-2)=1 for x>2 if x-2<0 (x<2), then |x-2|=-(x-2)=-(x-2) so |x-2|/{x-2}=-(x-2)/(x-2)=-1 for x<2 since we are looking to the right of the 2, then we are looking at values near 2 that are greater than 2 so the limit is 1

OpenStudy (anonymous):

ohhok..gotcha..thank u soo much!!

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