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

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.

• Have to tried wolfram alpha to evaluate it? Apr 7 '19 at 4:36
• No. But I personally believe that the limit doesn't exist. Apr 7 '19 at 4:40
• @mathmaniac you cant say the limit DNE because it only approaches form one side. It can be $\pm \infty$, but cannot be DNE Apr 7 '19 at 4:41
• Yeah good point @Aniruddh Venkatesan. But what about $\lim\limits_{x \rightarrow 0^+} \sin \left ( \frac 1 x \right )$? Does the limit is $\pm \infty$? Apr 7 '19 at 4:43
• @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 $f(b_n) \to -\infty$. Apr 7 '19 at 4:54

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.

• 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 Apr 7 '19 at 5:46
• 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 $0$. Apr 7 '19 at 5:48
• Verified: tinyurl.com/y4pktnpj <--- a_n tinyurl.com/y48vga57 <--- b_n For b_n you need to use "approximate form" Apr 7 '19 at 5:56

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: $$x=\frac1{\pi n}, n\in N \Rightarrow L=0\\ x\ne \frac1{\pi n}, n\in N \Rightarrow L=1.$$