Questions on Logic: Finite and infinite sets answered by real tutors!

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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(x)e^(sin(pi/x))
x->0+
: 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(x)e^(sin(pi/x))
x->0+

Answer by bucky(1732) About Me  (Show Source):
You can put this solution on YOUR website!
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.