How do you find the limit of #(3/x)^(1/x)# as x approaches infinity?

Question

Luke Alvarez · Accepted Answer

#y=1# #(3/oo)^(1/oo)=0^0#-> this is an indeterminate form so we use l'Hopitals' Rule Let #y=(3/x)^(1/x)# #lny=ln(3/x)^(1/x)# #lny=1/x ln(3/x)# #lny=ln(3/x)/x# #lim_(x->oo) lny = lim_(x->oo) ln(3/x)/x # #lim_(x->oo) lny =lim_(x->oo)((1/(3/x))*-3/x^2)/1# #lim_(x->oo)lny=lim_(x->oo)(x/3 xx-3/x^2)# #lim_(x->oo)lny=lim_(x->oo)-1/x# #lny=-1/oo# #lny=0# #e^0=y# #y=1#

Isabella Ackerman · Answer

undefined To find the limit of (3/x)^(1/x) as x approaches infinity, we can rewrite the expression using the properties of logarithms. Taking the natural logarithm of both sides, we get ln((3/x)^(1/x)). Using the logarithmic property ln(a^b) = b * ln(a), we can rewrite this as (1/x) * ln(3/x). Simplifying further, we have ln(3/x)/x.

Now, we can apply the limit properties. As x approaches infinity, 3/x approaches 0, and ln(3/x) approaches ln(0), which is undefined. However, we can rewrite ln(3/x) as ln(3) - ln(x) using the logarithmic property ln(a/b) = ln(a) - ln(b).

So, the expression becomes (ln(3) - ln(x))/x. As x approaches infinity, ln(x) also approaches infinity. Therefore, we have (ln(3) - infinity)/infinity, which is an indeterminate form.

To evaluate this indeterminate form, we can apply L'Hôpital's rule. Taking the derivative of the numerator and denominator separately, we get (-1/x)/(1). Simplifying, we have -1/x.

Now, as x approaches infinity, -1/x approaches 0. Therefore, the limit of (3/x)^(1/x) as x approaches infinity is 1.