Question #2ca19

1 Answer
Oct 18, 2016

#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#.