# How do you find the limit of e^sqrtx/(sqrt(e^x+1) as x->oo?

Apr 6, 2018

${\lim}_{x \to \infty} {e}^{\sqrt{x}} / \sqrt{{e}^{x} + 1} = 0$.

#### Explanation:

Let $L = {\lim}_{x \to \infty} {e}^{\sqrt{x}} / \sqrt{{e}^{x} + 1}$.

Since we have a square root, it'd be easier to calculate the limit of the function squared. In other words,

${L}^{2} = {\lim}_{x \to \infty} {\left({e}^{\sqrt{x}} / \sqrt{{e}^{x} + 1}\right)}^{2} = {\lim}_{x \to \infty} {e}^{2 \sqrt{x}} / \left({e}^{x} + 1\right)$

Let's rewrite this as the following:

L^2 = lim_(x->oo) e^(2sqrtx)/e^x * color(red)(e^x/(e^x+1)

If you do the multiplication, it's the same as before.

L^2 = lim_(x->oo) e^(2sqrtx-x) * color(red)(lim_(x->oo)e^x/(e^x+1)

The second limit is, intuitively, equal to $1$, and you can prove it by using $\text{L'Hôpital's rule}$.

${\lim}_{x \to \infty} {e}^{x} / \left({e}^{x} + 1\right) = \frac{\frac{d}{\mathrm{dx}} {e}^{x}}{\frac{d}{\mathrm{dx}} {e}^{x} + 1} = {e}^{x} / {e}^{x} = 1$.

Therefore,

${L}^{2} = {\lim}_{x \to \infty} {e}^{2 \sqrt{x} - x} = {e}^{{\lim}_{x \to \infty} 2 \sqrt{x} - x}$

${L}^{2} = {e}^{{\lim}_{x \to \infty} \sqrt{x} \left(2 - \sqrt{x}\right)} = {e}^{\infty \cdot \left(- \infty\right)} = {e}^{- \infty} = 0$

$\therefore L = 0$.