$$\lim_{x \to -∞}{\sqrt{x^2 -x} + x}.$$ First I rationalized, to get $$\frac{x^2-x-x^2}{\sqrt{x^2-x}-x}$$ Then I wanted to factor the whole thing by the largest power(x) in the denominator, to get: $$\frac{-x/x}{1/x(\sqrt{x^2-x}-x)}$$ After simplifying: $$\frac{-1}{\sqrt{1-1/x}-1}$$ And evaluating at negative infinity: $$\frac{-1}{1-0-1}$$ But this is incorrect as the limit would not exist. Where did I go wrong? Am I right in my procedure?
Thanks
$\endgroup$4 Answers
$\begingroup$Where did I go wrong?
The following is wrong :
$$\frac{\sqrt{x^2-x}}{x}=\sqrt{1-\frac 1x}$$
Note that for $x\lt 0$ $$\frac{\sqrt{x^2-x}}{x}=\frac{\sqrt{x^2(1-\frac 1x)}}{x}=\frac{\sqrt{x^2}\sqrt{1-\frac 1x}}{x}=\frac{|x|\sqrt{1-\frac 1x}}{x}=\frac{\color{red}{-}x\sqrt{1-\frac 1x}}{x}=\color{red}{-}\sqrt{1-\frac 1x}$$
$\endgroup$ $\begingroup$$$\lim _{ x\to -∞ }{ \sqrt { x^{ 2 }-x } +x } =\lim _{ x\to -∞ }{ \frac { \left( \sqrt { x^{ 2 }-x } +x \right) \left( \sqrt { x^{ 2 }-x } -x \right) }{ \left( \sqrt { x^{ 2 }-x } -x \right) } } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \sqrt { x^{ 2 }-x } -x \right) } } =\\ =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \left| x \right| \sqrt { 1-\frac { 1 }{ x } } -x \right) } } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( -x\sqrt { 1-\frac { 1 }{ x } } -x \right) } } =\lim _{ x\to -∞ }{ \frac { -x }{ -x\left( \sqrt { 1-\frac { 1 }{ x } } +1 \right) } } =\\ =\lim _{ x\to -∞ }{ \frac { 1 }{ \left( \sqrt { 1-\frac { 1 }{ x } } +1 \right) } } =\frac { 1 }{ 2 } $$
$\endgroup$ $\begingroup$Remember that $\sqrt{A^2}=|A|$.
we have that
$$(\forall x<0)\;\;\; \sqrt{x^2-x}+x=$$
$$|x|\sqrt{1-\frac{1}{x}}+x=$$
$$x(1-\sqrt{1-\frac{1}{x}})=$$
$$\frac{1}{1+\sqrt{1-\frac{1}{x}}}$$
and the limit when $x\to -\infty\;$ is $\frac{1}{2}$.
$\endgroup$ $\begingroup$Your limit can be rewritten as $$ \lim_{x\to +\infty}\sqrt{x^2+x}-x. $$ Using the fact that, as $x\to \infty$ $$ \sqrt{1+\frac{1}{x}}=1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right) $$ you conclude that $$ \sqrt{x^2+x}-x=x\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)-1\right) \to \frac{1}{2}. $$
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