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this my my simple question : lim x-0 (√x+2-√2/x) help any ideas
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is x under whole thing or just under sqrt[2]
yes is under whole
Multiply top and bottom by Sqrt[x+2]+Sqr[2]
Then you would be able to cancel x
1/sqrt[x+2]+Sqrt[2] 1/sqrt[2]+sqrt[2] 1/2(sqrt[2])
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infinite cayse lim=f(0) so u'll have sqr(0+2)-sqr(2/0) limsqr2=sqr2 and limsqr2/0=infinite sqr2-infinte=-infinite
\[\lim_{x \rightarrow 0} (\sqrt{x+2}-\sqrt{2}) / x = \lim (x+2-2)/ x(\sqrt{x+2} +\sqrt{2})= \lim 1/\sqrt{x+2} +\sqrt{2}=1/(2\sqrt{2})\]
thanks very much
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