Prove that `lim_(x->0^+)sqrt(x)e^(sin(pi/x)) = 0` .

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Reference: http://en.wikipedia.org/wiki/Squeeze_theorem

As noted above, the product rule cannot be used to find the limit of the given function because the limit of `e^(sin(pi/x))` does not exist.

But, the fact that the limit of `sqrt(x) e^(sin(pi/x))` is 0 can be proven using the Squeeze theorem, which says that if, for the functions f, g, and h such that

`f(x)<=g(x) <=h(x) ` in the vicinity of a

AND `lim_(x->a) f(x) = lim_(x->a) h(x) = L` (L being a finite number)

THEN `lim_(x->a) g(x)` also exists and equals L.

In our case,...

(The entire section contains 3 answers and 343 words.)

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