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

limits

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.

Best Answer

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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