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

In summary: Read moreIn summary, the conversation discusses the limit of a function involving the greatest integer function. The correct limit is found to be 1, but there is confusion about why it is not 2. It is explained that the greatest integer function is not continuous, so the limit cannot be found by swapping the order of operations. The function is graphed to better understand its behavior near 0, and it is determined that the limit is indeed 1.
  • #1
Ujjwal28
32
0

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...
 
Physics news on Phys.org
  • #2
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?
 
  • #3
Lol yes it the same question I too was shell shocked to see the answer
 
  • #4
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?
 
  • #5
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.
 
  • #6
How did you get to that? Could you please explain?
 
  • #7
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))]##?
 
  • Like
Likes SammyS
  • #8
I don't know what effect does it really have? So please explain it to me in detail
 
  • #9
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)) \ ? ##
 
  • Like
Likes Samy_A
  • #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?
 

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

What is a limit in calculus?

A limit in calculus is a fundamental concept that represents the value that a function approaches as its input approaches a certain value. It is the value that the function "approaches" or gets closer to, but may not necessarily reach at that specific input value.

How do you solve a limit?

To solve a limit, you can use algebraic manipulation, substitution, or graphical analysis. In this specific problem, substitution is used to plug in the value of x=1 into the given function and evaluate the resulting expression.

What is the difference between left and right-hand limits?

A left-hand limit is the value that a function approaches from the left side of the input value, while a right-hand limit is the value that a function approaches from the right side. In other words, a left-hand limit is evaluated by taking the values of x that are slightly smaller than the input value, while a right-hand limit is evaluated by taking the values of x that are slightly larger than the input value.

Can a limit exist even if the function is undefined at that point?

Yes, a limit can exist at a point where the function is undefined. This is because a limit represents the value that a function approaches, not the actual value at that point. For example, in this problem, the function is undefined at x=1, but the limit at x=1 is still solvable and has a finite value.

What does it mean if a limit does not exist?

If a limit does not exist, it means that the function does not approach a single value as the input approaches the specified point. This can happen if the function has a jump or a discontinuity at that point, or if the left and right-hand limits are not equal. In this case, the limit does exist and has a finite value.

Similar threads

Calculate the limit cos(x)/sin(x) when x approaches 0
Replies
12
Views
2K
Limit of x^α.sin²(x)/(x+1) as x approaches infinity is?
Replies
13
Views
2K
Limit of 0^0: Evaluating x^sinx
Replies
19
Views
2K
Finding the Maximum Value of f(x)=x(1-x)^n in [0,1]: A Calculus Problem
Replies
3
Views
1K
Find f(x,y) given partial derivative and initial condition
Replies
6
Views
1K
Finding the limit using a trig identity
Replies
5
Views
1K
Prove that this series converges uniformly on R
Replies
4
Views
999
Does x sin(1/x) exist as x approaches zero?
Replies
15
Views
6K
Proving Limit as x Approaches 0 for f(x)/x is 1?
Replies
7
Views
1K
Why lim x->0 e^-x - 1 / x = -1
Replies
7
Views
2K

Hot Threads

Recent Insights

Back
Top