SOLUTION: Of Calculus I: Prove that the limit as x approaches zero from the right of the function "The square root of x times e to the power of the sine of pi over x" is equal to zero. <

Algebra ->  Finite-and-infinite-sets -> SOLUTION: Of Calculus I: Prove that the limit as x approaches zero from the right of the function "The square root of x times e to the power of the sine of pi over x" is equal to zero. <      Log On


   

Question 85338: Of Calculus I:
Prove that the limit as x approaches zero from the right of the function
"The square root of x times e to the power of the sine of pi over x"
is equal to zero.


In mathematical terms, that is
Prove 
 lim sqrt%28x%29e%5E%28sin%28pi%2Fx%29%29
x->0+

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Find:
.
lim sqrt%28x%29e%5E%28sin%28pi%2Fx%29%29
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%28pi%2Fx%29. 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%5E%28sin%28pi%2Fx%29%29
.
has a finite value ranging from e%5E%28-1%29 to e%5E%28%2B1%29. Since this range of values
is finite (and positive) the limiting factor in this problem is sqrt%28x%29 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.