#lim_(xrarroo)(e^(2x)sin(1/x))/x^2#
Note that: #(e^(2x)sin(1/x))/x^2 = e^(2x)/x^3 * sin(1/x)/(1/x)#
Now, as #xrarroo#, the first ratio increases without bound, while the second goes to #1#.
#lim_(xrarroo)(e^(2x)sin(1/x))/x^2 = lim_(xrarroo)e^(2x)/x^3 * lim_(xrarroo)sin(1/x)/(1/x)#
# = oo#
Further Explanation
Here is the reasoning that led to the solution above.
#lim_(xrarroo)(e^(2x)sin(1/x))/x^2# has initial form #(oo*0)/oo#.
This is an indeterminate form, but we cannot apply l'Hospital's Rule to this form.
We could rewrite it as #(e^(2x))/(x^2/sin(1/x))# to get the form #oo/oo# to which we could apply l'Hospital. However, I don't particularly want to take the derivative of that denominator.
Recall that #lim_(thetararr0)sintheta/theta = 1#.
So that #lim_(xrarroo)sin(1/x)/(1/x) = 1#.
This is what motivates the rewriting used above.
#(e^(2x)sin(1/x))/x^2 = e^(2x)/x^3 * sin(1/x)/(1/x)#.
As #x# increases without bound, #e^x# goes to infinity much faster that #x^3# (faster than any power of #x#).
So, #e^(2x) = (e^x)^2# blows up even faster.
If you do not have this fact available, use l'Hospital's rule to get
#lim_(xrarroo)e^(2x)/x^3 = lim_(xrarroo)(2e^(2x))/(3x^2)#
# = lim_(xrarroo)(8e^(2x))/(6) = oo#
