Answer:
0
Step-by-step explanation:
Let us call
![f(x)=\frac{\sqrt[]{x}}{\csc x}](https://img.qammunity.org/2023/formulas/mathematics/college/x4h3g0fjeaafn44s9xtpylk3hrx24r687g.png)
The function is continuous on the interval [0, 2pi]; therefore,

To evaluate the limit itself, we use L'Hopital's rule which says

Now in our case, we have
![\lim _(n\to\pi)\frac{\sqrt[]{x}}{\csc x}=\lim _(n\to\pi)\frac{\frac{d\sqrt[]{x}}{dx}}{(d \csc x)/(dx)}](https://img.qammunity.org/2023/formulas/mathematics/college/zgkdrxyh7okrb6cncpk67p5wzuwgw3trso.png)
![=\lim _(n\to\pi)\frac{d\sqrt[]{x}}{dx}/(d\csc x)/(dx)](https://img.qammunity.org/2023/formulas/mathematics/college/pn981yszjvb1jzw4dwvbapnxta8d2vcn3z.png)
![=\frac{1}{2\sqrt[]{x}}/(-(\cos x)/(\sin^2x))](https://img.qammunity.org/2023/formulas/mathematics/college/y1tukpdlo0hfrkrz6oqke39xq12ab1yqdm.png)
since

Therefore, we have
![\lim _(n\to\pi)\frac{\sqrt[]{x}}{\csc x}=\lim _(n\to\pi)\frac{1}{2\sqrt[]{x}}/(-(\cos x)/(\sin^2x))](https://img.qammunity.org/2023/formulas/mathematics/college/1otbe2qy2lhg6u1di223h1b1u12my0q35g.png)
![=\lim _(n\to\pi)-\frac{1}{2\sqrt[]{x}}*(\sin^2x)/(\cos x)](https://img.qammunity.org/2023/formulas/mathematics/college/sv7cmwkmw2f0moj2158l6zd9zhar9h1od5.png)
Putting in x = π into the above expression gives
![-\frac{1}{2\sqrt[]{x}}*(\sin^2x)/(\cos x)\Rightarrow-\frac{1}{2\sqrt[]{\pi}}*(\sin^2\pi)/(\cos\pi)](https://img.qammunity.org/2023/formulas/mathematics/college/5kkppn4aflxt2fk34syj9333l232zyh3fv.png)

Hence,
![=\lim _(n\to\pi)-\frac{1}{2\sqrt[]{x}}*(\sin^2x)/(\cos x)=0](https://img.qammunity.org/2023/formulas/mathematics/college/mu6ciklry5beq9ionxh156lbcugzfi0g7k.png)
Therefore, we conclude that
![\boxed{\lim _(n\to\pi)\frac{\sqrt[]{x}}{\csc x}=0.}](https://img.qammunity.org/2023/formulas/mathematics/college/qftwb2oqoflrh2rn875m3ucerlcjb0uiis.png)
which is our answer!