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

1. doc.brown

$\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)}$

2. doc.brown

I've got 1-1 on the bottom, which is total garbage.

3. anonymous

Was the first one the original limit and the next few are your steps?

4. doc.brown

Yes

5. anonymous

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.

6. anonymous

consider $$x^2+2x-3=x^2+2x+1-4=(x+1)^2-2^2\\\qquad\qquad\quad=((x+1)-2)(x+3)\\\qquad\qquad\quad=(\sqrt{x+1}-\sqrt2)(\sqrt{x+1}+\sqrt2)(x+3)$$$$\frac{\sqrt{2}(\sqrt{x+1}-\sqrt{2})(x^2-9)}{x^2+2x-3}=\frac{\sqrt2(\sqrt{x+1}-\sqrt2)(x+3)(x-3)}{(\sqrt{x+1}-\sqrt2)(\sqrt{x+1}+\sqrt2)(x+3)}\\\qquad\qquad\qquad\qquad\qquad\qquad=\frac{\sqrt2(x-3)}{\sqrt{x+1}+\sqrt2}$$

7. anonymous

so as $$x\to1$$ we get $$\sqrt2(-2)/(2\sqrt2)=-1$$