# How do you find lim sqrt(x^2+1)-x as x->oo?

Jun 10, 2017

Let $\text{f} \left(x\right) = {\left({x}^{2} + 1\right)}^{\frac{1}{2}} - x$.

Then we can write,

$\text{f} \left(x\right) = {\left({x}^{2} \cdot \left(1 + \frac{1}{x} ^ 2\right)\right)}^{\frac{1}{2}} - x$
$\text{f} \left(x\right) = \left\mid x \right\mid \cdot {\left(1 + \frac{1}{x} ^ 2\right)}^{\frac{1}{2}} - x$.

Then, we can expand ${\left(1 + \frac{1}{x} ^ 2\right)}^{\frac{1}{2}}$ using the binomial theorem for $\left\mid \frac{1}{x} ^ 2 \right\mid < 1$.

$\left\mid \frac{1}{x} ^ 2 \right\mid < 1$ for $\frac{1}{x} ^ 2 < 1$ as $\frac{1}{x} ^ 2$ is always positive.
$\frac{1}{x} ^ 2 < 1$ for ${x}^{2} > 1$, so $x > 1$ or $x < - 1$, i.e. $\left\mid x \right\mid > 1$.

Then for $\left\mid x \right\mid > 1$,

$\text{f} \left(x\right) = \left\mid x \right\mid \cdot \left(1 + \left(\frac{1}{2}\right) \cdot \frac{1}{x} ^ 2 + \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \cdot \frac{1}{x} ^ 4 + \ldots\right) - x$
$\text{f} \left(x\right) = \left\mid x \right\mid - x + \frac{1}{2} \cdot \frac{\left\mid x \right\mid}{x} ^ 2 + \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \cdot \frac{\left\mid x \right\mid}{x} ^ 4 + \ldots$

Clearly ${\lim}_{x \setminus \to \setminus \infty} \frac{\left\mid x \right\mid}{x} ^ n = 0$ for $n > 1$.

Then, as this is expansion is valid for any $x > 1$,

${\lim}_{x \setminus \to \setminus \infty} \text{f} \left(x\right) = \left\mid x \right\mid - x$.

$\left\mid x \right\mid = \left\{\begin{matrix}x & x \ge 0 \\ - x & x \le 0\end{matrix}\right.$.

Then, we conclude,

${\lim}_{x \setminus \to + \setminus \infty} \text{f} \left(x\right) = 0$,
${\lim}_{x \setminus \to - \setminus \infty} \text{f} \left(x\right) = + \setminus \infty$.

Jun 10, 2017

${\lim}_{x \rightarrow \infty} \left(\sqrt{{x}^{2} + 1} - x\right)$ takes indeterminate form $\infty - \infty$

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

$= \frac{{x}^{2} + 1 - {x}^{2}}{\sqrt{{x}^{2} + 1} + x}$

$= \frac{1}{\sqrt{{x}^{2} + 1} + x}$

Now use
$\sqrt{{x}^{2} + 1} = \sqrt{{x}^{2}} \sqrt{1 + \frac{1}{x} ^ 2}$ for all $x \ne 0$

$= \left\mid x \right\mid \sqrt{1 + \frac{1}{x} ^ 2}$ for all $x \ne 0$

$= x \sqrt{1 + \frac{1}{x} ^ 2}$ for all $x > 0$ (We want ${\lim}_{x \rightarrow \infty}$)

Returning to

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

$= \frac{1}{x \sqrt{1 + \frac{1}{x} ^ 2}}$

As $x \rightarrow \infty$, we get a limit of $0$

Ar $x \rightarrow - \infty$, the expression $\sqrt{{x}^{2} + 1} - x$ takes non-indeterminate form $\infty - \left(- \infty\right) = \infty + \infty = \infty$