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

[Ujjwal28]

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...

 

[Tom MS]

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?

 

[Ujjwal28]

Lol yes it the same question I too was shell shocked to see the answer

 

[Samy_A]

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?

 

[SammyS]

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.

 

[Ujjwal28]

How did you get to that? Could you please explain?

 

[Samy_A]

How did you get to that? Could you please explain?

Can you explain why you think that $\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)] = [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]$?

 

[Ujjwal28]

I don't know what effect does it really have? So please explain it to me in detail

 

[Samy_A]

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.

 

[Ujjwal28]

I have no idea what you're telling me

 

[Samy_A]

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.

 

[Ujjwal28]

Someone please explain it to me

 

[SammyS]

Someone please explain it to me

Under what conditions is $\ \lim_{x\to a} f(g(x)) = f(\lim_{x\to a} g(x)) \ ?$

 

[Ujjwal28]

I'm unaware about it please tell me

 

[SammyS]

Under what conditions is $\ \lim_{x\to a} f(g(x)) = f(\lim_{x\to a} g(x)) \ ?$

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 $\displaystyle \ \left[ \frac{\sin(x-1)}{x-1} \right ] \ .$

 

[Mark44]

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$\rfloor$

 

[SammyS]

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 \$

 

[Ray Vickson]

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 $h$.

 

[ehild]

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)

 

[Ujjwal28]

I don't know how to procede

 

[Ray Vickson]

I don't know how to procede

What is stopping you from taking the advice offered in post #19?

 

[ehild]

I don't know how to procede

What is the greatest integer of 1.999999?