Question 85338
Find:
.
lim {{{sqrt(x)e^(sin(pi/x))}}}
x->0+
.
Since nobody else has taken a shot at this, maybe I can give you a way of looking at it.
.
Look at the exponent of e and recognize the limits on {{{sin(pi/x)}}}. The sine function
is limited to values between -1 and +1.  So the exponent of e will range in value from
-1 to +1.  That means that the term:
.
{{{e^(sin(pi/x))}}}
.
has a finite value ranging from {{{e^(-1)}}} to {{{e^(+1)}}}. Since this range of values
is finite (and positive) the limiting factor in this problem is {{{sqrt(x)}}} and since
x is approaching 0+, the expression gets closer and closer to zero times the finite value
of the term involving e.  Therefore, the limit is zero, just as the problem proposes that
you prove.
.
Hope this gets you on the right track and at least gives you a feel for a way of proving the
premise.