How to solve $\lim\limits_{x\to 0}\sin^{\sqrt{2x}}(x)$?

Question

$$\lim\limits_{x\to 0}\sin^{\sqrt{2x}}(x)$$

The link to the original image.

I don't know how to solve this limitation.The only idea I have is to separate $\sqrt{x}$ and $\sqrt{2}$ in the exponent and not sure what to do next. Thank you in advance.

Answer 1

As $x \to 0, \sin x \sim x$, so you can write $$ (\sin x)^{\sqrt{2x}} \sim x^{\sqrt{2x}} = e^{\sqrt{2x}\log x} = e^{\frac{\log x}{\frac{1}{\sqrt{2x}}}} $$ Now you can use the continuity of $\exp$ and use L'Hospital\s to get $$ e^{-\sqrt{8x}} $$ So the limit is $1$.

Answer 2

hint

When $ x $ goes to $ 0 $ from the right,

$\sin(x)>0 $ and

$$\sin^{\sqrt{2x}}(x)=e^{\sqrt{2x}\ln(\sin(x))}$$

Now, use $$\ln(\sin(x))= \ln(x) + \ln(\frac{\sin(x)}{x})$$ and

$$\sqrt{2x}\ln(x)= 2\sqrt{2}\sqrt{x}\ln(\sqrt{x})$$

the limit is $ =e^0=1$.

Answer 3

Technically speaking $\lim_{x\to0}\sin^{\sqrt{2x}}(x)$ does not exist, because $\sin^{\sqrt{2x}}(x)$ is not defined for $x\lt0$. One can, however, hope to evaluate the one-sided limit as $x\to0$ from the right. We have

$$\lim_{x\to0^+}\sin^{\sqrt{2x}}(x)=\lim_{x\to0^+}\left(\left(\sin(x)\over x\right)^{\sqrt2x}x^{\sqrt{2x}}\right)=\left(\lim_{x\to0^+}\left(\sin(x)\over x \right)^{\sqrt{2x}} \right)\left(\lim_{x\to0^+}x^{\sqrt{2x}} \right)\\=1^0\cdot\left(\lim_{x\to0^+}x^{\sqrt{2x}} \right)$$

using the familiar limit ${\sin(x)\over x}\to1$ as $x\to0$, so it remains to evaluate $\lim_{x\to0^+}x^{\sqrt{2x}}$. For this it's convenient to let $x=u^2$ with $u\gt0$, so that

$$\lim_{x\to0^+}x^{\sqrt{2x}}=\lim_{u\to0^+}(u^u)^{2\sqrt2}=1^{2\sqrt2}=1$$

using the familiar limit $u^u\to1$ as $u\to0^+$. (If either of these "familiar" limits is not familiar, that's a separate question.) In all, we find

$$\lim_{x\to0^+}\sin^{\sqrt{2x}}(x)=1$$

Remark: The original version of the question gave only an image asking for "$\lim_{x\to0}\sin^{\sqrt{2x}}$," i.e., lacking any argument for the sine function. The OP might have asked about that as well.