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Mathematics 13 Online
OpenStudy (anonymous):

evaluate the limit algebraically (attached)

OpenStudy (anonymous):

\[\lim_{x \rightarrow 2+} \frac{ x+3 }{ 2 }\]

OpenStudy (anonymous):

plug 2, gives (2+3)/2 = 2.5

OpenStudy (anonymous):

oh... hahah didnt realize it was that easy. thanks.

OpenStudy (anonymous):

can you show me how to do another one?

OpenStudy (anonymous):

link?

OpenStudy (anonymous):

\[\lim_{y \rightarrow 0-} \frac{ \left| y \right| }{ y }\] i dont understand how the answer is -1

OpenStudy (anonymous):

ok, 0- means it goes from left and it approaches 0

OpenStudy (anonymous):

so 0- is a veryyyyyyy small number, and since its at 0's left, it is a negative number

OpenStudy (anonymous):

something like -0.00000000000000000000000001

OpenStudy (anonymous):

now put that into the equation and see what u get

OpenStudy (anonymous):

-1 thanks!

OpenStudy (anonymous):

np ;)

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