lim x--->1 (x-1)/[(x^2 + x)^1/2 - (2x)^1/2] I'm stuck on this question
T tobalax New member Joined Sep 26, 2006 Messages 1 Sep 26, 2006 #1 lim x--->1 (x-1)/[(x^2 + x)^1/2 - (2x)^1/2] I'm stuck on this question
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 27, 2006 #2 Multiply top and bottom by the conjugate of the denominator. \(\displaystyle \L\\\frac{(x-1)}{\sqrt{x^{2}+x}-\sqrt{2x}}\cdot\frac{\sqrt{x^{2}+x}+\sqrt{2x}}{\sqrt{x^{2}+x}+\sqrt{2x}}\) =\(\displaystyle \L\\\frac{\sout{(x-1)}(\sqrt{x^{2}+x}+\sqrt{2x}}{x\sout{(x-1)}}\) You have: \(\displaystyle \L\\\lim_{x\to\1}\frac{\sqrt{x^{2}+x}+\sqrt{2x}}{x}\)
Multiply top and bottom by the conjugate of the denominator. \(\displaystyle \L\\\frac{(x-1)}{\sqrt{x^{2}+x}-\sqrt{2x}}\cdot\frac{\sqrt{x^{2}+x}+\sqrt{2x}}{\sqrt{x^{2}+x}+\sqrt{2x}}\) =\(\displaystyle \L\\\frac{\sout{(x-1)}(\sqrt{x^{2}+x}+\sqrt{2x}}{x\sout{(x-1)}}\) You have: \(\displaystyle \L\\\lim_{x\to\1}\frac{\sqrt{x^{2}+x}+\sqrt{2x}}{x}\)
