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Evaluate

$$\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin \left ( {\frac {1} {x}} \right ) + \sqrt x \sin \left ( {\frac {1} {x}} \right ) \right )^x.$$

I tried by taking log but it wouldn't work because there are infinitely many points in any neighbourhood of $0$ where $\ln \left ( \sin {\frac {1} {x}} \right )$ doesn't exist. How to overcome this situation?

Any help will be highly appreciated.Thank you very much for your valuable time.

EDIT $:$ I have observed that if the above limit exists at all when $x \rightarrow 0^+$ then the limit of $$\frac {\left ( 2 \sin \left ( {\frac {1} {x}} \right ) + \sqrt x \sin \left ( {\frac {1} {x}} \right ) \right )^x} {\left (2 + \sqrt x \right )^x}$$ as $x \rightarrow 0^{+}$ also exists since $$\lim\limits_{x \rightarrow 0^+} \left (2 + \sqrt x \right )^x = 1 \neq 0.$$ Therefore if the above limit will exist then the limit $$\lim\limits_{x \rightarrow 0^+} \left ( \sin \left ( \frac 1 x \right ) \right )^x$$ will also exist.

Which definitely doesn't exist.

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  • $\begingroup$ Have to tried wolfram alpha to evaluate it? $\endgroup$ – Juniven Apr 7 at 4:36
  • $\begingroup$ No. But I personally believe that the limit doesn't exist. $\endgroup$ – math maniac. Apr 7 at 4:40
  • $\begingroup$ @mathmaniac you cant say the limit DNE because it only approaches form one side. It can be $\pm \infty$, but cannot be DNE $\endgroup$ – Aniruddh Venkatesan Apr 7 at 4:41
  • $\begingroup$ Yeah good point @Aniruddh Venkatesan. But what about $\lim\limits_{x \rightarrow 0^+} \sin \left ( \frac 1 x \right )$? Does the limit is $\pm \infty$? $\endgroup$ – math maniac. Apr 7 at 4:43
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    $\begingroup$ @AniruddhVenkatesan Strictly speaking, that's not true. One sided limits can also fail even after including $\pm \infty$: e.g. $f(x) = (1/x)\sin(1 / x)$ does not have a one-sided limit as $x \to 0^+$. You have at least two sequences $a_n =((\pi / 2) + 2\pi n)^{-1}$ and $b_n = ((3\pi / 2) + 2\pi n)^{-1}$, both approaching $0$ from the right such that $f(a_n) \to +\infty$ and $f(b_n) \to -\infty$. $\endgroup$ – 0XLR Apr 7 at 4:54
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Good reasoning with factorizing the expression: $$\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin \left ( {\frac {1} {x}} \right ) + \sqrt x \sin \left ( {\frac {1} {x}} \right ) \right )^x=\\ \lim\limits_{x \rightarrow 0^+} \left ( 2 + \sqrt x \right )^x\cdot \lim\limits_{x \rightarrow 0^+} \left((\sin \left ( {\frac {1} {x}}\right)\right)^x.$$ But the second limit does not exist: $$x=\frac1{\pi n}, n\in N \Rightarrow L=0\\ x\ne \frac1{\pi n}, n\in N \Rightarrow L=1.$$

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Let $f(x) = \big(2\sin(\frac{1}{x}) + \sqrt{x}\sin(\frac{1}{x})\big)^x$ where $x > 0$ and $\sin(\frac{1}{x}) \geq 0$ (the reason for this domain in the update below). $\lim\limits_{x \to 0^+} f(x)$ does not exist because you can construct two sequences, both of which approach $0$ from the right but $f$ evaluated on the sequences approach two different limits. Consider the first sequence: $$a_n = \frac{1}{\pi + n\pi} \quad n \in \mathbb{N}$$ Then, for any $n \in \mathbb{Z}_+$, $\sin(\pi + n\pi) = 0$. So: $$f(a_n) = \Big(2\sin(\pi + n\pi) + \sqrt{\frac{1}{\pi + n\pi}}\sin(\pi + n\pi)\Big)^{\frac{1}{\pi + n\pi}} = 0^\frac{1}{\pi + n\pi} = 0$$ So $\lim\limits_{n \to \infty}f(a_n) = \lim\limits_{n \to \infty}0 = 0$. Next consider $$b_n = \frac{1}{\frac{\pi}{2} + 2\pi n} \quad n \in \mathbb{N}$$ Then for any $n \in \mathbb{Z}_+$, $\sin(\frac{\pi}{2} + 2\pi n) = 1$. So: $$f(b_n) = \Big(2\sin(\frac{\pi}{2} + 2\pi n) + \sqrt{\frac{1}{\frac{\pi}{2} + 2\pi n}}\sin(\frac{\pi}{2} + 2\pi n)\Big)^{\frac{1}{\frac{\pi}{2} + 2\pi n}} = \Big(2 + \sqrt{b_n}\Big)^{b_n}$$ As $n \to \infty$, $b_n \to 0$ and $f(b_n) = \Big(2 + \sqrt{b_n}\Big)^{b_n} \to \Big(2 + \sqrt{0}\Big)^{0} = 1$.

DOMAIN EXPLANATION:

$f(x) = \big(2\sin(\frac{1}{x}) + \sqrt{x}\sin(\frac{1}{x})\big)^x$ is not defined for all real $x > 0$. For example, plug $x_0 = \frac{2}{3\pi}$. Then, $\sin(\frac{1}{x_0}) = \sin(\frac{3\pi}{2}) = -1$ and $f(x_0) = (-1)^\frac{2}{3\pi}(2 + \sqrt{\frac{2}{3\pi}})^\frac{2}{3\pi}$. But $(-1)^\frac{2}{3\pi}$ does not make sense:

Indeed, an irrational exponent of a negative number does not in general refer to single value; it is a multi-valued expression and to make matters worse, the values are complex, not real.

In fact, continuing in this vein, $f(x)$ is undefined arbitrarily close to $0$. Just consider the sequence $$x_n = \frac{1}{\frac{3\pi}{2} + 2\pi n}$$ which clearly approaches $0$ from the right and $$f(x_n) = (-1)^\frac{1}{\frac{3\pi}{2} + 2\pi n}\Big(2 + \sqrt{\frac{1}{\frac{3\pi}{2} + 2\pi n}}\Big)^\frac{1}{\frac{3\pi}{2} + 2\pi n}$$ But $(-1)^\frac{1}{\frac{3\pi}{2} + 2\pi n}$ again does not make any sense.

Thus, a valid domain for $f(x)$ must prevent $f(x)$ from producing such complex values. One choice would be to pick all real $x > 0$ such that $\sin(\frac{1}{x}) \geq 0$. This requires us to delete all open intervals of the form $(\frac{1}{2\pi k}, \frac{1}{2\pi k - \pi}),\ k \in \mathbb{Z}_+$ from $\mathbb{R}_+$ to get this domain of $f(x)$: $$D = \mathbb{R}_+ - \bigcup_{k \in \mathbb{Z}_+}\big(\frac{1}{2\pi k}, \frac{1}{2\pi k - \pi}\big)$$ Fortunately, the sequences $a_n$ and $b_n$ used above remain within the domain so that the analysis goes through with the caveat that a slightly more general notion of a limit is being used:

If $f : D \subseteq \mathbb{R} \to \mathbb{R}$ is a function, $x_0 \in \mathbb{R}$ is a limit point of $D$ and $L \in \mathbb{R}$, then we say $L$ is the limit of $f$ as $x \in D$ approaches $x_0$ if for all $\epsilon > 0$, there is $\delta > 0$ such that $$f\big(D \cap (x_0 - \delta, x_0 + \delta) - \{x_0\}\big) \subseteq (L - \epsilon, L + \epsilon)$$ Also, to approach $x_0$ from the right, we merely change the condition to $$f\big(D \cap (x_0, x_0 + \delta)\big) \subseteq (L - \epsilon, L + \epsilon)$$

In particular, in this notion of limits, $f$ can be undefined arbitrarily close to $x_0$ i.e. we do not need an open interval around $x_0$ on which $f$ is defined everywhere. As long as every open interval around $x_0$ contains some point $x \neq x_0$ on which $f$ is defined, this notion of limits work.

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  • $\begingroup$ Doesn't $f(a_n)$ have $0^\frac{1}{\pi+n\pi}=0^0$? Also, my little computer program shows f(a_n) converges to 1 $\endgroup$ – Gareth Ma Apr 7 at 5:46
  • $\begingroup$ I showed that each $f(a_n)$ is equal to $0$ even before you take a limit. So the sequence $f(a_1), f(a_2), f(a_3), \ldots$ is just a sequence of $0$'s converging to $0$. $\endgroup$ – 0XLR Apr 7 at 5:48
  • $\begingroup$ Verified: tinyurl.com/y4pktnpj <--- a_n tinyurl.com/y48vga57 <--- b_n For b_n you need to use "approximate form" $\endgroup$ – Gareth Ma Apr 7 at 5:56

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