# What are some other ways of solving $y=\lim_{x \to 0^+} (2\sin(\sqrt x) + \sqrt x\sin{1\over x})^x$?

I have come across many limit questions of the form $$0^0$$

For instance , here is one example

$$y=\lim_{x \to 0^+} (2\sin(\sqrt x) + \sqrt x\sin{1\over x})^x$$

In this example if I take logarithm on both sides , it becomes $$\ln y = \lim_{x \to 0^+} x\ln (2\sin(\sqrt x) + \sqrt x\sin{1\over x})$$

And then I make the argument that as $$x \to 0$$ ,$$x$$ reaches zero more dominantly than $$\ln (2\sin(\sqrt x) + \sqrt x\sin{1\over x})$$ reaches $$-\infty$$ . And so $$\ln y = 0$$ , thus , $$y=1$$

But I realised tht this the argument I make in almost all such cases , and it got to me thinking are all limits of the form $$0^0$$ eventually turn out to be 1 ?

So what I am looking for is :

(a) Various other ways of solving the above limit (as many different ways are possible) , but please avoid using L'hopital's rule as it would be too tedious to work with here.

(b) Counter examples , that is , limits of the form $$0^0$$ which are not equal to $$1$$

Edit : I got the plenty counter examples , now I am only looking for other ways of solving the above limit

• For any real number $a \neq 0$ we have $(\frac 1 n)^{\ln (1/a) /ln n} \to a$. – Kavi Rama Murthy Jul 2 at 6:00
• No, the indeterminate form does not always come out to be $1$. And that is in particular true when you consider multi variable calculus, where more than not the answer won't be $1$ – imranfat Jul 2 at 6:02

$$0^0$$ we call "Uncertainty", because if we take different sequences with limits $$0$$ then we obtain different results: $$\begin{array} {} \left ( \frac{1}{n} \right )^{\frac{1}{n}}=e^{\frac{1}{n} \ln \frac{1}{n}} \to 1 \\ \left ( \frac{1}{n} \right )^{e^{-n^2}} = e^{\frac{1}{n} \ln e^{-n^2}} \to 0 \\ \left ( \frac{1}{n} \right )^{e^{-n}} = e^{\frac{1}{n} \ln e^{-n}} \to e^{-1} \end{array}$$ last line makes clear, that we can obtain any finite number $$a>0$$. One casuistic example we obtain if allow negative base (somebody forbid complex numbers?) $$\left ( -\frac{1}{n} \right )^{e^{-n^2}} = e^{-\frac{1}{n} \ln e^{-n^2}} \to +\infty$$

and last requirement $$\lim_{x \to 0^+} (2\sin(\sqrt x) + \sqrt x\sin{1\over x})^x = \lim_{x \to 0^+}e^{x \ln (2\sin(\sqrt x) + \sqrt x\sin{1\over x})}$$ now $$x \ln (2\sin(\sqrt x) + \sqrt x\sin{1\over x})=x \ln(2\sin(\sqrt x))+x\ln \left (1+\frac{\sqrt x\sin{1\over x}}{2\sin(\sqrt x)} \right ) \to0$$

$$\lim\limits_{x \to 0} \left( 2x \right)^{\frac{3}{5 - \ln(4x)}} = \left[ 0^0 \right] = \lim\limits_{x \to 0} e^{\frac{3 \cdot \ln(2x)}{5 - \ln(4x)}} = \left[ \text{L'Hopital} \right] = \lim\limits_{x \to 0} e^{\frac{\frac{d}{dx}(3 \cdot \ln(2x))}{\frac{d}{dx}(5 - \ln(4x))}} = \lim\limits_{x \to 0} e^{\frac{\frac{3}{x}}{\frac{-1}{x}}} = \boxed{e^{-3}}$$

• Yeah , I got the counter example ,thanks , but I am still looking for other ways of solving the above limit (that's why I am not green checking either of the two answers. – ARROW Jul 2 at 6:44

No, the limit of the form 0^0 not always comes out to be 1

Here is a counter example,

Here is an approach which takes care of the oscillating term $$\sin(1/x)$$.
Clearly we have $$|\sin(1/x)|\leq 1$$ and therefore $$2\sin\sqrt{x}-\sqrt{x}\leq 2\sin\sqrt{x}+\sqrt {x} \sin\frac {1}{x}\leq 2\sin\sqrt{x}+\sqrt{x}$$ Raising each term to power $$x$$ we get an inequality of the form $$g(x) \leq f(x) \leq h(x)$$ where $$f(x)$$ is the function under limit in question. You can now show that $$g(x) \to 1,h(x)\to 1$$ so that by squeeze theorem $$f(x) \to 1$$.
We have $$g(x) =(2\sin\sqrt{x}-\sqrt{x})^x=x^{x/2}\left(2\cdot\frac{\sin\sqrt{x}}{\sqrt{x}}-1\right)^x$$ The expression $$x^{x/2}$$ tends to $$1$$ (prove this via the standard limit that $$x\log x\to 0$$) and the expression in large parentheses tends to $$1$$ so that the second factor also tends to $$1$$. It follows that $$g(x) \to 1$$ and in a similar manner $$h(x) \to 1$$.
Although some authors define $$0^0=1$$, does not mean that $$\lim_{(x,y)\to(0,0)}x^y=1\tag1$$ There are many paths to $$(0,0)$$ that do not lead to $$1$$. For example, if $$x\gt0$$, $$x^{\frac\lambda{\log(x)}}=e^\lambda\tag2$$ Taking the limit as $$x\to0^+$$ obviously gives $$e^\lambda$$.
However, for $$0\le x\le\frac{\pi^2}{16}$$, the concavity of $$\sin(x)$$ gives $$\frac{2\sqrt2}\pi\sqrt{x}\le\sin\left(\sqrt{x}\right)$$. Therefore, $$\left(\frac{4\sqrt2}\pi-1\right)\sqrt{x}\le2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\tfrac1x\right)\le3\sqrt{x}\tag3$$ Furthermore, for $$a\gt0$$, \begin{align} \lim_{x\to0^+}\left(a\sqrt{x}\right)^{\,x} &=\lim_{x\to0^+}\left(\sqrt2\,a\right)^x\lim_{x\to0^+}(x/2)^{x/2}\\[3pt] &=1\cdot1\tag4 \end{align} The Squeeze Theorem, with $$(3)$$ and $$(4)$$, shows that $$\lim_{x\to0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\tfrac1x\right)\right)^x=1\tag5$$