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doc.brown
 one year ago
The limit as x > 1
doc.brown
 one year ago
The limit as x > 1

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doc.brown
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\sqrt{2}(\sqrt{x+1}\sqrt{2})(x^29)}{x^2+2x3}\]\[\frac{\sqrt{2}(\sqrt{x+1}\sqrt{2})(x+3)(x3)}{(x+3)(x1)}\]\[\frac{\sqrt{2}(\sqrt{x+1}\sqrt{2})(x3)}{(x1)}\]

doc.brown
 one year ago
Best ResponseYou've already chosen the best response.0I've got 11 on the bottom, which is total garbage.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Was the first one the original limit and the next few are your steps?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. You may have gotten 0 again in the denominator, but you also got 0 in the numerator, so you can still work with it. Try multiplying by \[\frac{\sqrt{x+1}+\sqrt{2}}{\sqrt{x+1}+\sqrt{2}}\]and see if anything cancels.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider \(x^2+2x3=x^2+2x+14=(x+1)^22^2\\\qquad\qquad\quad=((x+1)2)(x+3)\\\qquad\qquad\quad=(\sqrt{x+1}\sqrt2)(\sqrt{x+1}+\sqrt2)(x+3)\)$$\frac{\sqrt{2}(\sqrt{x+1}\sqrt{2})(x^29)}{x^2+2x3}=\frac{\sqrt2(\sqrt{x+1}\sqrt2)(x+3)(x3)}{(\sqrt{x+1}\sqrt2)(\sqrt{x+1}+\sqrt2)(x+3)}\\\qquad\qquad\qquad\qquad\qquad\qquad=\frac{\sqrt2(x3)}{\sqrt{x+1}+\sqrt2}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so as \(x\to1\) we get \(\sqrt2(2)/(2\sqrt2)=1\)
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