Question

Find the one-sided limit (if it exists). (If the limit does not exist, enter DNE.)

Answer 1

0

Step-by-step explanation:

Let us call

`f(x)=\frac{\sqrt[]{x}}{\csc x}`

The function is continuous on the interval [0, 2pi]; therefore,

`\lim _(x\to\pi^+)f(x)=\lim _(x\to\pi^-)f(x)`

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

`\lim _(x\to c)(a(x))/(b(x))=\lim _(x\to c)(a^(\prime)(x))/(b^(\prime)(x))`

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)}`

`=\lim _(n\to\pi)\frac{d\sqrt[]{x}}{dx}/(d\csc x)/(dx)`

`=\frac{1}{2\sqrt[]{x}}/(-(\cos x)/(\sin^2x))`

since

`(d\csc x)/(dx)=-(\cos x)/(\sin^2x)`

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))`
`=\lim _(n\to\pi)-\frac{1}{2\sqrt[]{x}}*(\sin^2x)/(\cos x)`

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)`
`=0`

Hence,

`=\lim _(n\to\pi)-\frac{1}{2\sqrt[]{x}}*(\sin^2x)/(\cos x)=0`

Therefore, we conclude that

`\boxed{\lim _(n\to\pi)\frac{\sqrt[]{x}}{\csc x}=0.}`

which is our answer!