Let's shuffle things around so that there's only one exponential:
#lim_(x->oo)(3^x)/(x2^x)=lim_(x->oo)(3//2)^x/x->oo/oo#
The above representation of the limit has both the numerator and denominator growing without bound as #x# increases to infinity. The question is: does one of them grow faster than the other? Or perhaps, does the ratio of their growing speeds approach a finite value?
We can answer this question by finding the rates that both sides of the fraction approach infinity, and equating the original limit to the limit of the ratio of these rates. In other words, we are going to find their derivatives.
#color(white)(lim_(x->oo)(3^x)/(x2^x))=lim_(x->oo)(d/dx(3//2)^x)/(d/dxx)#
This is called L'Hôpital's Rule, and you can use it whenever you have a fractional limit that approaches #0/0# or #+-oo/oo.#
Continuing, we have
#color(white)(lim_(x->oo)(3^x)/(x2^x))=lim_(x->oo)((3/2)^xln(3/2))/(1)#
The multiplier #ln(3//2)# is just a constant, so we can pull it outside:
#color(white)(lim_(x->oo)(3^x)/(x2^x))=ln(3/2)lim_(x->oo)(3/2)^x#
Now, what's left inside our limit? It's an exponential function of #x#, with a base that's greater than #1.# And for any base #b>1,# we know #lim_(x->oo)b^x=oo.#
So we get
#color(white)(lim_(x->oo)(3^x)/(x2^x))=ln(3/2)timesoo#
#color(white)(lim_(x->oo)(3^x)/(x2^x))=oo#
Therefore,
#lim_(x->oo)(3^x)/(x2^x)=oo.#
Bonus:
In general, exponential functions (like #b^x#) always grow to infinity faster than power functions (like #x^n#), as long as #b>1.# As #x->oo,# #b^x# gains more multiples of #b,# while the number of #x"'s"# in #x^n# remains fixed at #n.# No matter how large #n# is, there is always a finite value of #x# at which any exponential growth #b^x# ("infinite multiplication") starts to exceed #x^n# ("finite multiplication").
