Question

lim x goes to positive infinity sqrt(x^2 + 3x) - sqrt(x^2 + x)

Answer 1

\[\lim_{x\to\infty}\left(\sqrt{x^2+3x}-\sqrt{x^2+x}\right)\cdot\frac{\sqrt{x^2+3x}+\sqrt{x^2+x}}{\sqrt{x^2+3x}+\sqrt{x^2+x}}\]
Then you have
\[\lim_{x\to\infty}\frac{(x^2+3x)-(x^2+x)}{\sqrt{x^2+3x}+\sqrt{x^2+x}}=\lim_{x\to\infty}\frac{2x}{\sqrt{x^2+3x}+\sqrt{x^2+x}}\]
Factor \(\sqrt{x^2}\) from the denominator:
\[\lim_{x\to\infty}\frac{2x}{\sqrt{x^2}\left(\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x}}\right)}\]
By definition, you have \(\sqrt{x^2}=|x|=x\), since \(|x|=x\) for positive \(x\). This is because \(x\to\infty\). So, you have
\[\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x}}}\]