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Let the real-valued function

$$f(x)=\begin{cases} \left|{\sin\frac{\pi}{2x}}\right|^{x},& x>0\, \text{ and } x\neq\frac{1}{2n}, \;n\in\mathbb{N}\\ 1,& x=\frac{1}{2n},\; n\in\mathbb{N}\;. \end{cases}$$

Find, if it exists, $\displaystyle\mathop{\lim}\limits_{x\rightarrow{0^{+}}}{f(x)}$.

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For each $n$, find $x_n \in (\frac{1}{2n}, \frac{1}{n})$ so that $|\sin(\frac{\pi}{2x_n})|^{\frac{1}{2n}} < 1/2$.

This is possible as $\sin(\frac{\pi}{2x})$ has a zero at $\frac{1}{2n}$.

Therefore, $|\sin(\frac{\pi}{2x_n})|^{x_n} \le |\sin(\frac{\pi}{2x_n})|^{\frac{1}{2n}} < 1/2$.

Since $\lim_{n\to\infty}f(\frac{1}{2n}) = 1$, and $f(x_n) < 1/2$ for all $n$, $\lim_{x\to 0^+} f(x)$ doesn't exist.

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I beat this limit to arrive at it equaling this $$\exp\left\{\lim_{x\to\infty} {1\over x}\log\left|\sin\left({\pi\over 2}x\right)\right|\right\}.$$ You can see that this last limit is hopeless. The $1/x$ factor is not going to damp out the horrendous behavior that occurs because of the zeroes of the sine function.

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In fact you made the substitution $t=\frac{1}{x}$. But now you must reverse also $x\neq\frac{1}{2n}$, $n\in\mathbb{N}$. So, must be $t\neq2n$, for $n\in\mathbb{N}$. Exactly at these points the $\sin\big({t\,\tfrac{\pi}{2}}\bigr)$ is zero. Also $\log1=0$. So, you must examine the $\displaystyle\mathop{\lim}\limits_{t\rightarrow{+\infty}}{g(t)}$ of the function $g(t)=\left\{{\begin{array}{lc} \dfrac{1}{t}\,\log\bigr|{\sin\big({t\,\tfrac{\pi}{2}}\bigr)}\bigr|\,, & 0<t \ and \ t\neq2n\,, \ n\in\mathbb{N}\\ 0\,, & t=2n\,, \ n\in\mathbb{N} \end{array}}\right.$ – Grigorios Kostakos Feb 25 '12 at 2:04
Yes. But even weeding those points out leaves and intractible mess. – ncmathsadist Feb 25 '12 at 3:49

No, the limit does not exist. Let us consider two extreme cases:

  • When $\frac{\pi}{2x}$ is of the form $n \pi$, then $\sin (\frac{\pi}{2x}) = 0$ so the expression is equal to zero.

  • When $\frac{\pi}{2x}$ is of the form $n \pi + \frac{\pi}{2}$, then $\sin (\frac{\pi}{2x})=1$ so the expression is equal to one.

Notice that the function $|\sin (\frac{\pi}{2x})|^x$ is continuous. So even though the function $f(x)$ is defined to be 1 in these two extreme cases, we can get arbitrarily close to 0 in the first case -- that is, by setting $x$ extremely close to but not exactly equal to $\frac{1}{4n}$.

Since we're alternating between 1 and something arbitrarily close to 0, there is no limit as $x$ approaches 0.

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