The limit as x -> 1

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The limit as x -> 1

Mathematics
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\[\frac{\sqrt{2}(\sqrt{x+1}-\sqrt{2})(x^2-9)}{x^2+2x-3}\]\[\frac{\sqrt{2}(\sqrt{x+1}-\sqrt{2})(x+3)(x-3)}{(x+3)(x-1)}\]\[\frac{\sqrt{2}(\sqrt{x+1}-\sqrt{2})(x-3)}{(x-1)}\]
I've got 1-1 on the bottom, which is total garbage.
Was the first one the original limit and the next few are your steps?

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Yes
Ok. 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.
consider \(x^2+2x-3=x^2+2x+1-4=(x+1)^2-2^2\\\qquad\qquad\quad=((x+1)-2)(x+3)\\\qquad\qquad\quad=(\sqrt{x+1}-\sqrt2)(\sqrt{x+1}+\sqrt2)(x+3)\)$$\frac{\sqrt{2}(\sqrt{x+1}-\sqrt{2})(x^2-9)}{x^2+2x-3}=\frac{\sqrt2(\sqrt{x+1}-\sqrt2)(x+3)(x-3)}{(\sqrt{x+1}-\sqrt2)(\sqrt{x+1}+\sqrt2)(x+3)}\\\qquad\qquad\qquad\qquad\qquad\qquad=\frac{\sqrt2(x-3)}{\sqrt{x+1}+\sqrt2}$$
so as \(x\to1\) we get \(\sqrt2(-2)/(2\sqrt2)=-1\)

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