Question

Question #2ca19

Answer 1

`lim_(x->oo)g(x) = 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->oo)g(x) = lim_(x->oo)x^(1/sqrt(x))`

`=lim_(x->oo)e^ln(x^(1/sqrt(x)))`

`=lim_(x->oo)e^(1/sqrt(x)ln(x))`

`=e^(lim_(x->oo)ln(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 `oo/oo` indeterminate form, we will apply L'Hopital's rule.

`lim_(x->oo)ln(x)/sqrt(x) = lim_(x->oo)(d/dxln(x))/(d/dxsqrt(x))`

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

`=lim_(x->oo)2/sqrt(x)`

`=0`

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

`lim_(x->oo)g(x) =e^(lim_(x->oo)ln(x)/sqrt(x))`

`=e^0`

`=1`

So `g(x) -> 1` as `x->+oo`, meaning we have a horizontal asymptote at `y=1`.