Limits,Continuity:How to proceed when a combination of limit and GIF is given?

MathGeek19

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\(\displaystyle \displaystyle \mbox{If }\, L\, =\, \lim_{x\, \rightarrow\, 0}\, \left[\dfrac{\sin(x)}{x}\right],\, M\, =\, \lim_{x\, \rightarrow\, 0}\, \left[\dfrac{x^2}{\sin(x)}\right],\, N\, =\, \left[\lim_{x\, \rightarrow\, 0}\,\dfrac{\sin^2(x)}{x}\right],\, \mbox{ and }\, O\, =\, \left[\lim_{x\, \rightarrow\, 0}\,\dfrac{x^2}{\sin(x)}\right]\)

\(\displaystyle \mbox{(where [.] denotes the Greatest Integer Function, or GIF), then L+M+N+O =?

. . .(1) 0 . . .(2) 1 . . .(3) 2 . . .(4) 3 or 4
I understand that N=O=0. What is confusing me is L and M.
Here, if I put the limits first in each of L,M,N and O then I might get the correct answer but how do I determine whether to use the limit first or The GIF first?(As far as I remember the limit can be taken outside as well inside the GIF)
GIF = Greatest Integer Function.

The correct answer is option (1).\)
 
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\(\displaystyle \displaystyle \mbox{If }\, L\, =\, \lim_{x\, \rightarrow\, 0}\, \left[\dfrac{\sin(x)}{x}\right],\, M\, =\, \lim_{x\, \rightarrow\, 0}\, \left[\dfrac{x^2}{\sin(x)}\right],\, N\, =\, \left[\lim_{x\, \rightarrow\, 0}\,\dfrac{\sin^2(x)}{x}\right],\, \mbox{ and }\, O\, =\, \left[\lim_{x\, \rightarrow\, 0}\,\dfrac{x^2}{\sin(x)}\right]\)

\(\displaystyle \mbox{(where [.] denotes the Greatest Integer Function, or GIF), then }\, L\, +\, M\, +\, N\, +\, O\, \mbox{ is equal to:}\)

. . .(1) 0 . . .(2) 1 . . .(3) 2 . . .(4) 3 or 4

Here, if I put the limits first in each of L,M,N and O then I might get the correct answer but how do I determine whether to use the limit first or The GIF first?(As far as I remember the limit can be taken outside as well inside the GIF)

The correct answer is option (1).
First, besides being a graphics-file extension I have never seen GIF used in mathematics. What does it mean?

Do you understand the product theorem for limits? If so then you know that

\(\displaystyle \displaystyle{ {\lim _{x \to 0}}\frac{{{{\sin }^2}(x)}}{x}\, =\, {\lim _{x \to 0}}x \cdot {\left( {\frac{{\sin (x)}}{x}} \right)^2} \,=\, {\lim _{x \to 0}}x \cdot {\lim _{x \to 0}}{\left( {\frac{{\sin (x)}}{x}} \right)^2}\, =\, \left( 0 \right)\left( 1 \right) \,=\, 0}\)

So the correct answer is indeed 1.
 
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\(\displaystyle \displaystyle \mbox{If }\, L\, =\, \lim_{x\, \rightarrow\, 0}\, \left[\dfrac{\sin(x)}{x}\right],\, M\, =\, \lim_{x\, \rightarrow\, 0}\, \left[\dfrac{x^2}{\sin(x)}\right],\, N\, =\, \left[\lim_{x\, \rightarrow\, 0}\,\dfrac{\sin^2(x)}{x}\right],\, \mbox{ and }\, O\, =\, \left[\lim_{x\, \rightarrow\, 0}\,\dfrac{x^2}{\sin(x)}\right]\)

\(\displaystyle \mbox{(where [.] denotes the Greatest Integer Function, or GIF), then }\, L\, +\, M\, +\, N\, +\, O\, \mbox{ is equal to:}\)

. . .(1) 0 . . .(2) 1 . . .(3) 2 . . .(4) 3 or 4

Here, if I put the limits first in each of L,M,N and O then I might get the correct answer but how do I determine whether to use the limit first or The GIF first?(As far as I remember the limit can be taken outside as well inside the GIF)

The correct answer is option (1).
You should know that N=O=0 so that really only leaves L and M.

You can't take the limit inside the GIF. The GIF is defined as:
[x] = the largest integer n less than or equal to x, i.e. \(\displaystyle x\, -\, 1\, \lt\, n\, \le\, x\) so [-1.1]=-2. Thus what does that say about the continuity of [\(\displaystyle \frac{sin^2(x)}{x}\)] at x equal to zero and, because of that, the limit of that as x goes to zero? Or, in other words, are you sure that the limits, at least for [\(\displaystyle \frac{sin^2(x)}{x}\)] isn't as x is approached from the right?
 
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First, besides being a graphics-file extension I have never seen GIF used in mathematics. What does it mean?

Do you understand the product theorem for limits? If so then you know that

\(\displaystyle \displaystyle{ {\lim _{x \to 0}}\frac{{{{\sin }^2}(x)}}{x}\, =\, {\lim _{x \to 0}}x \cdot {\left( {\frac{{\sin (x)}}{x}} \right)^2} \,=\, {\lim _{x \to 0}}x \cdot {\lim _{x \to 0}}{\left( {\frac{{\sin (x)}}{x}} \right)^2}\, =\, \left( 0 \right)\left( 1 \right) \,=\, 0}\)

So the correct answer is indeed 1.
Apologies for causing confusion,I'm new here and I just copied the question and pasted it here.I have updated the question to make it more clear.
(I'm required to find the sum of L,M,N and O).
GIF stands for Greatest Integer Function here.
Yes,I do understand the product theorem but what is causing me confusion is the different combinations of limits inside and outside the GIF.
 
You should know that N=O=0 so that really only leaves L and M.

You can't take the limit inside the GIF. The GIF is defined as:
[x] = the largest integer n less than or equal to x, i.e. \(\displaystyle x\, -\, 1\, \lt\, n\, \le\, x\) so [-1.1]=-2. Thus what does that say about the continuity of [\(\displaystyle \frac{sin^2(x)}{x}\)] at x equal to zero and, because of that, the limit of that as x goes to zero? Or, in other words, are you sure that the limits, at least for [\(\displaystyle \frac{sin^2(x)}{x}\)] isn't as x is approached from the right?
I understand that N=O=0.
But L and M are confusing.
I am clear about the definition of GIF and the fact that it would be continuous at all values except of course integral values.
I am still confused, could you just explain just L and M?
Thanks.
 
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