Question

Evaluate the limit

Answer 1

\[\lim_{x \rightarrow 0^+}(\sin x)(\left[ \frac{ 1 }{ \sin x } \right]+\left[ \frac{ 2 }{ \sin x } \right])\]

Answer 2

[t] denotes the greatest integer less than or equal to t.

Answer 3

Just multiply and add

Answer 4

yeah ^^

Answer 5

@drawar, so the Floor function?

Answer 6

Yup

Answer 7

Multiply? How?

Answer 8

We'll wait for floors and celling ;)

Answer 9

\[\text{Sin}[x]\left(\frac{1}{\text{Sin}[x]}+\frac{2}{\text{Sin}[x]}\right)=3 \]

Answer 10

@drawar multiply as in distribute

Answer 11

\[\lim_{x\to0^+}\sin x\left(\left\lfloor\frac{1}{\sin x}\right\rfloor+\left\lfloor\frac{2}{\sin x}\right\rfloor\right)\]
Normal multiplicative distribution won't work here... @robtobey has the right answer, but I'm not sure how one would determine the limit analytically.

Answer 12

^ second that

Answer 13

Actually I have already worked out the answer myself but I'm not so sure about it, since an (unofficial) suggested solution offers a different approach.

Answer 14

Refer to a screen capture from Mathematica.

Answer 15

Essentially they're not so different, both make use of the well-known Squeeze Theorem but I derive a different bounds from the solution's.

Answer 16

@robtobey : The limit is indeed 3 anyway but it's not the answer that matters.

Answer 17

I got messed up, the \(sin(\frac{\pi}{2})\) could make up 1 and -1, so
\[\lim_{x \rightarrow 0^+}(\sin \frac{\pi}{2})(\left[ \frac{ 1 }{ \sin \frac{\pi}{2} } \right]+\left[ \frac{ 2 }{ \sin \frac{\pi}{2} } \right])\]
Since its approaches to positive infinity then it would give 3

Answer 18

postitive zero*

Answer 19

Oh I think I got it alr, I've just overlooked the fact that x is approaching 0 from the right-side, my solution cannot be completed w/o that.

Answer 20

Refer to the attachment for the application of L' Hôpital's rule

Answer 21

Thank you for the medal.

Answer 22

@robtobey Well, that doesn't make any sense, since [...] here denotes the ceiling function, not some kinds of square brackets. Thanks for taking the time on this question anyway! :)