# Homework Help: Limits and derivatives

1. May 20, 2016

### Ujjwal28

1. The problem statement, all variables and given/known data
What will be lim[2sin(x-1)/(x-1)], where x tends to 1?
[ ] denotes greatest integer function.
2. Relevant equations
Can I directly solve it using the formula sinx/x =1 when x tends to 0

3. 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...

2. May 20, 2016

### Tom MS

You really should be right. Can you make sure you wrote it down correctly?

3. May 20, 2016

### Ujjwal28

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

4. May 20, 2016

### Samy_A

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 continous function?

5. May 20, 2016

### SammyS

Staff Emeritus
This limit is not 2 .

The given answer, that the limit is 1, is indeed correct.

6. May 20, 2016

### Ujjwal28

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

7. May 20, 2016

### Samy_A

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

8. May 20, 2016

### Ujjwal28

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

9. May 20, 2016

### Samy_A

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. May 20, 2016

### Ujjwal28

I have no idea what you're telling me

11. May 20, 2016

### Samy_A

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. May 20, 2016

### Ujjwal28

Someone please explain it to me

13. May 20, 2016

### SammyS

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

14. May 20, 2016

### Ujjwal28

15. May 20, 2016

### SammyS

Staff Emeritus
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. May 20, 2016

### Staff: Mentor

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. May 20, 2016

### SammyS

Staff Emeritus
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: May 20, 2016
18. May 20, 2016

### Ray Vickson

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. May 20, 2016

### ehild

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: May 21, 2016
20. May 20, 2016

### Ujjwal28

I don't know how to procede

21. May 20, 2016

### Ray Vickson

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

22. May 21, 2016

### ehild

What is the greatest integer of 1.999999?