Limits and Derivatives: Solving lim[2sin(x-1)/(x-1)] as x approaches 1

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression lim[2sin(x-1)/(x-1)] as x approaches 1, where the brackets denote the greatest integer function. Participants are exploring the implications of applying the limit to a function that includes the greatest integer operation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss whether the limit can be evaluated directly using the known limit of sin(x)/x as x approaches 0. There are attempts to justify why the limit should yield 2, while others question the continuity of the greatest integer function at x=1. Some participants express confusion about the reasoning behind the limit's value and seek clarification on the implications of the greatest integer function in this context.

Discussion Status

The conversation is ongoing, with various interpretations being explored. Some participants have provided guidance on the continuity of the greatest integer function and its effect on the limit, while others are still grappling with the reasoning and implications of their approaches.

Contextual Notes

There is a recurring emphasis on the need to evaluate the behavior of the function near x=1, rather than at x=1 itself, as well as the importance of understanding the conditions under which certain limit theorems apply.

Ujjwal28
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Homework Statement


What will be lim[2sin(x-1)/(x-1)], where x tends to 1?
[ ] denotes greatest integer function.

Homework Equations


Can I directly solve it using the formula sinx/x =1 when x tends to 0

The Attempt at a Solution


Okay so the quantity inside [ ] can be written as ——>>2 sin(x-1)/(x-1).. And sin(x-1)/(x-1) should equAte to 1 and only 2 should be left inside the greAtest integer function and therefore the answer should be 2... But the answer is 1 what am I doing wrong please explain...
 
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Ujjwal28 said:

Homework Statement


What will be lim[2sin(x-1)/(x-1)], where x tends to 1?
[ ] denotes greatest integer function.

Homework Equations


Can I directly solve it using the formula sinx/x =1 when x tends to 0

The Attempt at a Solution


Okay so the quantity inside [ ] can be written as ——>>2 sin(x-1)/(x-1).. And sin(x-1)/(x-1) should equAte to 1 and only 2 should be left inside the greAtest integer function and therefore the answer should be 2... But the answer is 1 what am I doing wrong please explain...
You really should be right. Can you make sure you wrote it down correctly?
 
Lol yes it the same question I too was shell shocked to see the answer
 
Ujjwal28 said:

Homework Statement


What will be lim[2sin(x-1)/(x-1)], where x tends to 1?
[ ] denotes greatest integer function.

Homework Equations


Can I directly solve it using the formula sinx/x =1 when x tends to 0

The Attempt at a Solution


Okay so the quantity inside [ ] can be written as ——>>2 sin(x-1)/(x-1).. And sin(x-1)/(x-1) should equAte to 1 and only 2 should be left inside the greAtest integer function and therefore the answer should be 2... But the answer is 1 what am I doing wrong please explain...
What you claim is that ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)] = [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##.

But is that claim correct? Is [] a continuous function?
 
Ujjwal28 said:

Homework Statement


What will be lim[2sin(x-1)/(x-1)], where x tends to 1?
[ ] denotes greatest integer function.

Homework Equations


Can I directly solve it using the formula sinx/x =1 when x tends to 0

The Attempt at a Solution


Okay so the quantity inside [ ] can be written as ——>>2 sin(x-1)/(x-1).. And sin(x-1)/(x-1) should equAte to 1 and only 2 should be left inside the greAtest integer function and therefore the answer should be 2... But the answer is 1 what am I doing wrong please explain...
This limit is not 2 .

The given answer, that the limit is 1, is indeed correct.
 
How did you get to that? Could you please explain?
 
Ujjwal28 said:
How did you get to that? Could you please explain?
Can you explain why you think that ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)] = [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##?
 
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I don't know what effect does it really have? So please explain it to me in detail
 
Ujjwal28 said:
I don't know what effect does it really have? So please explain it to me in detail
Let's start with the RHS: ##\displaystyle \ [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##.
As you stated, ##\displaystyle \lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))=2##, so that ##\displaystyle \ [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]=[2]=2##.

For the LHS, ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)]##, you can't simply reverse the order of operations, swapping the limit with the []. That only works for continuous functions.

[] is not continuous at x=1. If that is not clear to you, try to prove this first.

To evaluate ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)]##, you first have to evaluate ##[2\sin(x-1)/(x-1)]## for x close to 1, but not equal to 1. Try it, and see what you get.
 
  • #10
I have no idea what you're telling me
 
  • #11
Ujjwal28 said:
I have no idea what you're telling me
That was quick.

Anyway, then we are back at post #7. Why did you think that the limit should be 2? Give your detailled reasoning, then someone will hopefully be able to help you find the error(s).
Because, as @SammyS wrote, the correct limit is 1.
 
  • #12
Someone please explain it to me
 
  • #13
Ujjwal28 said:
Someone please explain it to me
Under what conditions is ##\ \lim_{x\to a} f(g(x)) = f(\lim_{x\to a} g(x)) \ ? ##
 
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  • #14
I'm unaware about it please tell me
 
  • #15
SammyS said:
Under what conditions is ##\ \lim_{x\to a} f(g(x)) = f(\lim_{x\to a} g(x)) \ ? ##
Ujjwal28 said:
I'm unaware about it please tell me

It's dangerous to use a theorem without paying attention to the conditions under which the theorem is valid.

We can get back to that later. You might continue to ponder that.

As an alternate way to understand this:
Graph the function ##\displaystyle \ \left[ \frac{\sin(x-1)}{x-1} \right ] \ .##
 
  • #16
SammyS said:
It's dangerous to use a theorem without paying attention to the conditions under which the theorem is valid.

We can get back to that later. You might continue to ponder that.

As an alternate way to understand this:
Graph the function ##\displaystyle \ \left[ \frac{\sin(x-1)}{x-1} \right ] \ .##
Just to be clear, the function in question is ## \lfloor \frac{\sin(x-1)}{x-1}\rfloor##, the greatest integer function -- the largest integer less than or equal to ##\frac{\sin(x - 1)}{x - 1}##

The LaTeX for this is ##\lfloor ##\dots## \rfloor##
 
  • #17
Mark44 said:
Just to be clear, the function in question is ## \lfloor \frac{\sin(x-1)}{x-1}\rfloor##, the greatest integer function -- the largest integer less than or equal to ##\frac{\sin(x - 1)}{x - 1}##

The LaTeX for this is ##\lfloor ##\dots## \rfloor##
Right. Since OP was using [ ] , I continued using that.

With the standard floor symbols: ##\displaystyle \ \left\lfloor \frac{\sin(x-1)}{x-1} \right\rfloor \ ##
 
Last edited:
  • #18
Ujjwal28 said:
How did you get to that? Could you please explain?

Although the function ##f(h) = \sin(h)/h## has a limit of 1 as ##h \to 0## (where ##h = x-1##), the question is: how does ##f(h)## behave for ##h \neq 0## but near 0? Is it a little bit bigger than 1? Is it a little bit smaller than 1? You cannot answer these questions just from the limit alone; you need a bit more information.

If ##f(h) < 1## for small ##|h| \neq 0## then ##1 < 2 f(h) < 2 ## for small, nonzero values of ##h##, and that will mean that ##\lfloor 2f(h) \rfloor = 1## for small nonzero values of ##h##.
 
  • #19
Ujjwal28 said:
I don't know what effect does it really have? So please explain it to me in detail
Evaluate ##\lfloor 2 \frac {\sin(x-1)}{x-1}\rfloor## for x=0.9, 0.99, 0.999, and x= 1.1, 1.01, and 1.001. What do you get? So what is the limit if x-->1?
(Set your calculator to SCi digit 9)
 
Last edited:
  • #20
I don't know how to procede
 
  • #21
Ujjwal28 said:
I don't know how to procede

What is stopping you from taking the advice offered in post #19?
 
  • #22
Ujjwal28 said:
I don't know how to procede
What is the greatest integer of 1.999999?
 

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