# Question 2ca19

Oct 18, 2016

${\lim}_{x \to \infty} g \left(x\right) = 1$, giving a horizontal asymptote of $y = 1$

#### Explanation:

When trying to take the limit of an exponential function, we can convert it to an easier form by using logarithms.

${\lim}_{x \to \infty} g \left(x\right) = {\lim}_{x \to \infty} {x}^{\frac{1}{\sqrt{x}}}$

$= {\lim}_{x \to \infty} {e}^{\ln} \left({x}^{\frac{1}{\sqrt{x}}}\right)$

$= {\lim}_{x \to \infty} {e}^{\frac{1}{\sqrt{x}} \ln \left(x\right)}$

$= {e}^{{\lim}_{x \to \infty} \ln \frac{x}{\sqrt{x}}}$

where the last equality follows from the continuity of ${e}^{x}$.

Now we can evaluate the limit in the exponent and then substitute it back into the equation above. As a direct attempt at evaluating the limit produces an $\frac{\infty}{\infty}$ indeterminate form, we will apply L'Hopital's rule.

${\lim}_{x \to \infty} \ln \frac{x}{\sqrt{x}} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(x\right)}{\frac{d}{\mathrm{dx}} \sqrt{x}}$

=lim_(x->oo)(1/x)/(1/(2sqrt(x))#

$= {\lim}_{x \to \infty} \frac{2}{\sqrt{x}}$

$= 0$

Now that we have that limit, we can substitute it back into the exponent to get our result.

${\lim}_{x \to \infty} g \left(x\right) = {e}^{{\lim}_{x \to \infty} \ln \frac{x}{\sqrt{x}}}$

$= {e}^{0}$

$= 1$

So $g \left(x\right) \to 1$ as $x \to + \infty$, meaning we have a horizontal asymptote at $y = 1$.