Answer:
0
Explanation:
Let us call
[tex]f(x)=\frac{\sqrt[]{x}}{\csc x}[/tex]
The function is continuous on the interval [0, 2pi]; therefore,
[tex]\lim _{x\to\pi^+}f(x)=\lim _{x\to\pi^-}f(x)[/tex]
To evaluate the limit itself, we use L'Hopital's rule which says
[tex]\lim _{x\to c}\frac{a(x)}{b(x)}=\lim _{x\to c}\frac{a^{\prime}(x)}{b^{\prime}(x)}[/tex]
Now in our case, we have
[tex]\lim _{n\to\pi}\frac{\sqrt[]{x}}{\csc x}=\lim _{n\to\pi}\frac{\frac{d\sqrt[]{x}}{dx}}{\frac{d \csc x}{dx}}[/tex]
[tex]=\lim _{n\to\pi}\frac{d\sqrt[]{x}}{dx}\div\frac{d\csc x}{dx}[/tex]
[tex]=\frac{1}{2\sqrt[]{x}}\div(-\frac{\cos x}{\sin^2x})[/tex]
since
[tex]\frac{d\csc x}{dx}=-\frac{\cos x}{\sin^2x}[/tex]
Therefore, we have
[tex]\lim _{n\to\pi}\frac{\sqrt[]{x}}{\csc x}=\lim _{n\to\pi}\frac{1}{2\sqrt[]{x}}\div(-\frac{\cos x}{\sin^2x})[/tex][tex]=\lim _{n\to\pi}-\frac{1}{2\sqrt[]{x}}\times\frac{\sin^2x}{\cos x}[/tex]
Putting in x = π into the above expression gives
[tex]-\frac{1}{2\sqrt[]{x}}\times\frac{\sin^2x}{\cos x}\Rightarrow-\frac{1}{2\sqrt[]{\pi}}\times\frac{\sin^2\pi}{\cos\pi}[/tex][tex]=0[/tex]
Hence,
[tex]=\lim _{n\to\pi}-\frac{1}{2\sqrt[]{x}}\times\frac{\sin^2x}{\cos x}=0[/tex]
Therefore, we conclude that
[tex]\boxed{\lim _{n\to\pi}\frac{\sqrt[]{x}}{\csc x}=0.}[/tex]
which is our answer!
