E etotheipi New member Joined Apr 11, 2008 Messages 3 Apr 21, 2008 #1 Hi all. Im really stuck on this problem. I need to resolve the limits on this: \(\displaystyle \lim_{x \to \infty}x^{3/2}[(1+x+\frac{1}{x})^{1/2}-(1+x)^{1/2}]\) Any help will be greatly appreciated.
Hi all. Im really stuck on this problem. I need to resolve the limits on this: \(\displaystyle \lim_{x \to \infty}x^{3/2}[(1+x+\frac{1}{x})^{1/2}-(1+x)^{1/2}]\) Any help will be greatly appreciated.
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Apr 21, 2008 #2 \(\displaystyle \left( {\sqrt {1 + x + \frac{1}{x}} - \sqrt {1 + x} } \right)\frac{{\left( {\sqrt {1 + x + \frac{1}{x}} + \sqrt {1 + x} } \right)}}{{\left( {\sqrt {1 + x + \frac{1}{x}} + \sqrt {1 + x} } \right)}} = \frac{{\frac{1}{x}}}{{\left( {\sqrt {1 + x + \frac{1}{x}} + \sqrt {1 + x} } \right)}}\)
\(\displaystyle \left( {\sqrt {1 + x + \frac{1}{x}} - \sqrt {1 + x} } \right)\frac{{\left( {\sqrt {1 + x + \frac{1}{x}} + \sqrt {1 + x} } \right)}}{{\left( {\sqrt {1 + x + \frac{1}{x}} + \sqrt {1 + x} } \right)}} = \frac{{\frac{1}{x}}}{{\left( {\sqrt {1 + x + \frac{1}{x}} + \sqrt {1 + x} } \right)}}\)