1. ## Proving a limit.

Here is the question:

Prove that:

lim [x -> 0+] sqrt(x) e^sin(pi / x) = 0

(for clarification, it's a right-hand limit)

What's throwing me off here is the exponent of e, sin(pi / x) since that itself is undefined, and I'm unsure how that affects the question.

My friend has said that it's equal to zero because it is all values greater than zero, limited at zero.

How could I go about proving the initial limit statement?

2. Originally Posted by Tulki
Here is the question:

Prove that:

lim [x -> 0+] sqrt(x) e^sin(pi / x) = 0

(for clarification, it's a right-hand limit)

What's throwing me off here is the exponent of e, sin(pi / x) since that itself is undefined, and I'm unsure how that affects the question.

My friend has said that it's equal to zero because it is all values greater than zero, limited at zero.

How could I go about proving the initial limit statement?
Remember that

$\lim_{x \to a}\left[f(x)\cdot g(x)\right] = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(a)$.

So

$\lim_{x \to 0^+}\sqrt{x} e^{\sin{\frac{\pi}{x}}} = \lim_{x \to 0^+}\sqrt{x} \lim_{x \to 0+}e^{\sin{\frac{\pi}{x}}}$

$= 0 \lim_{x \to 0^+}e^{\sin{\frac{\pi}{x}}}$

$= 0$.

3. The largest value sin(pi/x) can have is 1 and it's smallest value is -1, right? So the largest value that $exp(sin(pi/x))$ can take is e, and it's smallest it can take is 1/e.

Now use the squeezing theorem.

4. Originally Posted by Tulki
Here is the question:

Prove that:

lim [x -> 0+] sqrt(x) e^sin(pi / x) = 0

(for clarification, it's a right-hand limit)

What's throwing me off here is the exponent of e, sin(pi / x) since that itself is undefined, and I'm unsure how that affects the question.

My friend has said that it's equal to zero because it is all values greater than zero, limited at zero.

How could I go about proving the initial limit statement?
Let $x = \frac{1}{t}$ and consider $\lim_{t \rightarrow + \infty} \frac{e^{\sin (\pi t) }}{\sqrt{t}}$. I suggest using the pinching (sandwich, squeeze) theorem.

5. Wow, speedy response!

Thanks, I forgot about the rule you stated initially.