What is the limit of  (sqrt(x^4 - 6x^2)) - x^2 as x goes to infinity?

Oct 19, 2015

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{4} - 6 {x}^{2}} - {x}^{2}\right) = - 3$

Explanation:

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{4} - 6 {x}^{2}} - {x}^{2}\right)$ has indeterminate form $\infty - \infty$.

So we'll do some algebra:

$\sqrt{{x}^{4} - 6 {x}^{2}} - {x}^{2} = \frac{\left(\sqrt{{x}^{4} - 6 {x}^{2}} - {x}^{2}\right)}{1} \cdot \frac{\left(\sqrt{{x}^{4} - 6 {x}^{2}} + {x}^{2}\right)}{\left(\sqrt{{x}^{4} - 6 {x}^{2}} + {x}^{2}\right)}$

$= \frac{\left({x}^{4} - 6 {x}^{2}\right) - {x}^{4}}{\sqrt{{x}^{4} - 6 {x}^{2}} + {x}^{2}}$ $\text{ }$ (has indeterminate form $\frac{\infty}{\infty}$)

$= \frac{- 6 {x}^{2}}{\sqrt{{x}^{4} \left(1 - \frac{6}{x} ^ 2\right)} + {x}^{2}}$ for $x \ne 0$

$= \frac{- 6 {x}^{2}}{\sqrt{{x}^{4}} \sqrt{\left(1 - \frac{6}{x} ^ 2\right)} + {x}^{2}}$ for $x \ne 0$

$= \frac{- 6 {x}^{2}}{{x}^{2} \left(\sqrt{1 - \frac{6}{x} ^ 2} + 1\right)}$ for $x \ne 0$

$= \frac{- 6}{\sqrt{1 - \frac{6}{x} ^ 2} + 1}$ for $x \ne 0$

Now as $x \rightarrow \infty$, we see that the ratio $\rightarrow \frac{- 6}{2}$

So, ${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{4} - 6 {x}^{2}} - {x}^{2}\right) = - 3$

Note that because the identity $\sqrt{{x}^{4}} = {x}^{2}$ is true for both positive and negative values of $x$, we also have

${\lim}_{x \rightarrow - \infty} \left(\sqrt{{x}^{4} - 6 {x}^{2}} - {x}^{2}\right) = - 3$