Question

lim x-> infinity ((sqrt(x^2-3x)-x) how do i find the limit? please help don't understand

Answer 1

\[\lim_{x\to\infty}\left(\sqrt{x^2-3x}-x\right)\\
\lim_{x\to\infty}\left(\sqrt{x^2}\sqrt{1-\frac{3}{x}}-x\right)\\
\lim_{x\to\infty}\left(|x|\sqrt{1-\frac{3}{x}}-x\right)\]
Since x is getting arbitrarily large (and positive), you know that |x| = x.
\[\lim_{x\to\infty}\left(x\sqrt{1-\frac{3}{x}}-x\right)\\
\lim_{x\to\infty}x\left(\sqrt{1-\frac{3}{x}}-1\right)\]
Evaluating directly yields the indeterminate form (infinity) times (zero). This can easily be fixed with some rewriting:
\[\lim_{x\to\infty}\frac{\sqrt{1-\frac{3}{x}}-1}{\frac{1}{x}}\]
Now, evaluating directly yields the indeterminate form 0/0. L'Hopital's rule should work from here.