# How do you find lim (sqrt(y+1)+sqrt(y-1))/y as y->oo using l'Hospital's Rule?

Nov 11, 2017

${\lim}_{y \to \infty} \frac{\sqrt{y + 1} + \sqrt{y - 1}}{y} = 0$

#### Explanation:

L'Hospital's Rule is applicable if in a limit ${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)}$, either as $x \to a$, $f \left(x\right)$ and $g \left(x\right)$ both tend to $0$ or to $\infty$. In such cases ${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = \frac{f ' \left(a\right)}{g ' \left(a\right)}$. If $\frac{f ' \left(a\right)}{g ' \left(a\right)} = \frac{\infty}{\infty}$ or $\frac{0}{0}$, then we go for $\frac{f ' ' \left(a\right)}{g ' ' \left(a\right)}$

Here we have $y$ and as $y \to \infty$, both numerator and denominator tend to $\infty$.

Hence ${\lim}_{y \to \infty} \frac{\sqrt{y + 1} + \sqrt{y - 1}}{y}$

= ${\lim}_{y \to \infty} \frac{\frac{1}{2 \sqrt{y + 1}} + \frac{1}{2 \sqrt{y - 1}}}{1}$

= $\frac{\frac{1}{2 \times \infty} + \frac{1}{2 \times \infty}}{1}$

= $0$