Question

Determine the limit when x goes to infinity sqrt(x^(2)+x)-x if it exists.

Answer 1

$The, initial, form, for, the, limit, is, indeterminate, \infty-\infty So, use, the, conjugate. \begin{array}{l} \left(\sqrt{x^{2}+x}-x\right)=\frac{\sqrt{x^{2}+x}-x}{1} \\ \frac{\sqrt{x^{2}+x}+x}{\sqrt{x^{2}+x}+x} \\ =\frac{x^{2}+x-x^{2}}{\sqrt{x^{2}+x}+x} \\ =\frac{x}{\sqrt{x^{2}+x}+x} \end{array} \lim _{x \rightarrow \infty} \frac{x}{\sqrt{x^{2}+x}+x} has, indeterminate, form, \frac{\infty}{\infty}, but, we, can, factor, and, reduce. We, know, that, \sqrt{x^{2}}=|x|, so, for, positive, x (which, is, all, we, are, concerned, about, for, a, limit, as, x, increases, without, bound,) we, have, \begin{array}{l} \frac{x}{\sqrt{x^{2}+x}+x}=\frac{x}{\sqrt{x^{2}} \sqrt{1+\frac{1}{x}+x}} \\ x \neq 0) \end{array} (for, all$$=\frac{x}{x \sqrt{1+\frac{1}{x}}+x} x>0) =\frac{x}{x\left(\sqrt{\left.1+\frac{1}{x}+1\right)}\right.} =\frac{1}{\sqrt{1+\frac{1}{x}}+1} \lim _{x \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac{1}{\sqrt{1}+1}=\frac{1}{2}$