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OpenStudy (anonymous):

this my my simple question : lim x-0 (√x+2-√2/x) help any ideas

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OpenStudy (anonymous):

is x under whole thing or just under sqrt[2]

OpenStudy (anonymous):

yes is under whole

OpenStudy (anonymous):

Multiply top and bottom by Sqrt[x+2]+Sqr[2]

OpenStudy (anonymous):

Then you would be able to cancel x

OpenStudy (anonymous):

1/sqrt[x+2]+Sqrt[2] 1/sqrt[2]+sqrt[2] 1/2(sqrt[2])

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OpenStudy (angela210793):

infinite cayse lim=f(0) so u'll have sqr(0+2)-sqr(2/0) limsqr2=sqr2 and limsqr2/0=infinite sqr2-infinte=-infinite

OpenStudy (anonymous):

\[\lim_{x \rightarrow 0} (\sqrt{x+2}-\sqrt{2}) / x = \lim (x+2-2)/ x(\sqrt{x+2} +\sqrt{2})= \lim 1/\sqrt{x+2} +\sqrt{2}=1/(2\sqrt{2})\]

OpenStudy (anonymous):

thanks very much

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