find lim sqr(x+9)-3/x x->0
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 5, 2010 #2 Multiply top and bottom by the conjugate, x+9+3\displaystyle \sqrt{x+9}+3x+9+3 Then it reduces down to something you can deal with. By the way, I assume you mean x+9−3x\displaystyle \frac{\sqrt{x+9}-3}{x}xx+9−3. You have x+9−3x\displaystyle \sqrt{x+9}-\frac{3}{x}x+9−x3 written. That is why grouping symbols are important.
Multiply top and bottom by the conjugate, x+9+3\displaystyle \sqrt{x+9}+3x+9+3 Then it reduces down to something you can deal with. By the way, I assume you mean x+9−3x\displaystyle \frac{\sqrt{x+9}-3}{x}xx+9−3. You have x+9−3x\displaystyle \sqrt{x+9}-\frac{3}{x}x+9−x3 written. That is why grouping symbols are important.
