# How do you evaluate [x/(x^2-d^2)^(1/2)]  as x approaches negative infinity?

Jul 11, 2016

-1

#### Explanation:

${\lim}_{x \to - \infty} \left[\frac{x}{{x}^{2} - {d}^{2}} ^ \left(\frac{1}{2}\right)\right]$

${\lim}_{x \to - \infty} s g n \left(x\right) \left[\frac{1}{1 - {\left(\frac{d}{x}\right)}^{2}} ^ \left(\frac{1}{2}\right)\right]$ because the denominator is always positive

and
${\lim}_{x \to - \infty} {\left(\frac{d}{x}\right)}^{2} = 0$

$\implies {\lim}_{x \to - \infty} \left[\frac{x}{{x}^{2} - {d}^{2}} ^ \left(\frac{1}{2}\right)\right] = - 1$

Jul 11, 2016

the reqd. limit$= 1.$

#### Explanation:

Let $f \left(x\right) = \left[\frac{x}{{x}^{2} - {d}^{2}} ^ \left(\frac{1}{2}\right)\right] = \left[\frac{x}{{x}^{2} \left(1 - {d}^{2} / {x}^{2}\right)} ^ \left(\frac{1}{2}\right)\right] = \left[\frac{x}{x \cdot {\left\{1 - {\left(\frac{d}{x}\right)}^{2}\right\}}^{\frac{1}{2}}}\right] = \left[\frac{\cancel{x}}{\cancel{x} \cdot {\left\{1 - {\left(\frac{d}{x}\right)}^{2}\right\}}^{\frac{1}{2}}}\right] = \left[\frac{1}{{\left\{1 - {\left(\frac{d}{x}\right)}^{2}\right\}}^{\frac{1}{2}}}\right]$

Now, letting, $x \rightarrow - \infty$, we have, $\frac{d}{x} \rightarrow 0$

Hence, the reqd. limit$= 1.$