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

Find the one-sided limit (if it exists). (If the limit does not exist, enter DNE.)-example-1

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Answer:

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!

User Pierre Ozoux
by
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