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

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### 3 Answers

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, `sin(pi/x)` , while undefined at x = 0, is between -1 and 1:

`-1<=sin(pi/x)<=1`

Since `e^x` is an increasing function, the following also has to be true:

`e^(-1) <=e^(sin(pi/x))<=e^1`

We can multiply all sides of this inequality by ` ` `sqrt(x)` , because it is always positive:

`sqrt(x) e^(-1)<=sqrt(x)e^(sin(pi/x)) <=sqrt(x)e^1`

This is true everywhere, including the vicinity of 0. So these three functions satisfy the conditions of the Squeeze theorem, because

`lim_(x->0+) sqrt(x)e^(-1) = lim_(x->0+) sqrt(x)e^1 = 0`

So the limit of the function "squeezed" in between is also 0:

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

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

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

First, the limit of a product is the product of the limits. So lim[f(x)g(x)]=LK where lim f(x)=L and lim g(x)=K, assuming the limits exist.

(1) limit as x approaches 0 from the right is 0.

(2) limit as x approaches zero of e^(sin(pi/x)) does not exist.

However, the value of e^(sin(pi/x)) lies between e^(-1) and e.

Since the limit of sqrt(x) exists and is 0, and e^(sin(pi/x)) is bounded, the limit is 0. (Zero times a finite number is 0.)

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