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Find the limit of [1-cos(x)]/x Watch

1. I am really struggling with this question. The hint is multiply top and bottom by 1 + cos(x) but that hasn't helped me so far
2. (Original post by Cggh90)
I am really struggling with this question. The hint is multiply top and bottom by 1 + cos(x) but that hasn't helped me so far
Hi,

Use the following identity

then multiply and divide the fraction x/2 and calculate
Note:
3. do you know what sin(x)/x tends to? do you know what sin(x) tends to? and do you know what 1+cos(x) tends to?
4. (Original post by kfkle)
do you know what sin(x)/x tends to? do you know what sin(x) tends to? and do you know what 1+cos(x) tends to?
sin(x)/x tends to 1 as x tends to 0.
sin(x) tends to 0

Does cos(x) + 1 tend to 2???

I've got to |sin(x)/x| |sin(x)/(1+cos(x)|

Stuck from here..!
5. (Original post by Cggh90)
Does cos(x) + 1 tend to 2???
The sum of continuous functions is continuous.
6. (Original post by IrrationalNumber)
The sum of continuous functions is continuous.
I don't really understand that but it does tend to 2 doesn't it?
7. (Original post by Cggh90)
I don't really understand that but it does tend to 2 doesn't it?
A continuous function is a function f for which as x tends to a f(x) tends to f(a). Cos is continuous. The function 1 is continuous. So the function f defined by f(x)=cos(x)+1 is continuous. That means that

In short, yes, and the reason is continuity.
8. Thanks Irrational, I need to take in all I can!

Ok so can you help me with this question.

We have [1-cos(x)]/x as x tends to 0.

[1-cos(x)]/x = 1-cos^2(x) / x(1+cos(x)

= sin^2(x) / x(1+cos(x) = sinx/x . sinx/1+cosx

Now I know sinx/x tends to 1 as x tends to 0
I know sinx tends to 0 as x tends to 0
and I know that 1+ cos(x) tends to 2 as x tends to 0.

What can I do with this?!
9. (Original post by Cggh90)
Thanks Irrational, I need to take in all I can!

Ok so can you help me with this question.

We have [1-cos(x)]/x as x tends to 0.

[1-cos(x)]/x = 1-cos^2(x) / x(1+cos(x)

= sin^2(x) / x(1+cos(x) = sinx/x . sinx/1+cosx

Now I know sinx/x tends to 1 as x tends to 0
I know sinx tends to 0 as x tends to 0
and I know that 1+ cos(x) tends to 2 as x tends to 0.

What can I do with this?!
Have a look at theorem 2.6 in your notes.
10. (Original post by kfkle)
Have a look at theorem 2.6 in your notes.
Does the squeeze rule come into play here?

Also on a side not, what did you get in your three class tests (excluding number theory as I don't do that). I just wondered as you seem like a stronger student to everyone I know!
11. (Original post by Cggh90)
Does the squeeze rule come into play here?

Also on a side not, what did you get in your three class tests (excluding number theory as I don't do that). I just wondered as you seem like a stronger student to everyone I know!
not the squeeze rule, no. seriously, go and look for theorem 2.6.

i got 9 in 1052, 20 in 1050, and 37 in 1048.

what marks did you get?
12. (Original post by kfkle)
not the squeeze rule, no. seriously, go and look for theorem 2.6.

i got 9 in 1052, 20 in 1050, and 37 in 1048.

what marks did you get?
Good work.

I got 8 in 1052, 20 in 1050 and 30 in 1048.

Yeh hard to believe I got 20/20 in this module but the exam was piss easy compared to the class sheets.

Do you average around the 80-90% in the courseworks?

And I'll look at theorem 2.6 now and try and make sense of it

EDIT: I don't know what sinx/1+cosx tends to .. ? I just know the two terms individually.. Wait give me one second lol
13. Which university are you both at?
14. For sin(x)/1+cos(x), let sin(x) = f(x) and 1+ cos(x) = g(x)

f(x) tends to 0 as x tends to 0 and 1+cos(x) tends to 0 as x tends to 0.

So sin(x)/1+cos(x) tends to 0/2 = 0

Then the same can be done with |sinx/x| . |sinx/1+cos(x) finding that it tends to 0 ...

So 1-cos(x)/x tends to 0 ???
15. (Original post by Cggh90)
Good work.

I got 8 in 1052, 20 in 1050 and 30 in 1048.

Yeh hard to believe I got 20/20 in this module but the exam was piss easy compared to the class sheets.

Do you average around the 80-90% in the courseworks?

And I'll look at theorem 2.6 now and try and make sense of it

EDIT: I don't know what sinx/1+cosx tends to .. ? I just know the two terms individually..
good job on the tests. averaging just over 90% on the assignments atm, how about you?

@IrrationalNumber: southampton university.
16. (Original post by IrrationalNumber)
Which university are you both at?
We don't actually know each other IRL, but Southampton university..

Where did yo/do you study?
17. (Original post by kfkle)
good job on the tests. averaging just over 90% on the assignments atm, how about you?

@IrrationalNumber: southampton university.
Are you one of the students who got 5 a's at a levels so came here due to the free tuition fees?

Anyway I'm doing alright.

I'm on 80% in LA, 70% in DE and 50% in this module. But my calculus class test boosted me up so I'm on decent grades.

I'd be delighted with a 2:1 come June 2013
18. (Original post by Cggh90)
Are you one of the students who got 5 a's at a levels so came here due to the free tuition fees?

Anyway I'm doing alright.

I'm on 80% in LA, 70% in DE and 50% in this module. But my calculus class test boosted me up so I'm on decent grades.

I'd be delighted with a 2:1 come June 2013
I do happen to have 5 A's yes, but that's not why I'm here. it's a long story.
19. (Original post by Cggh90)
We don't actually know each other IRL, but Southampton university..

Where did yo/do you study?
Warwick. Still there.
20. (Original post by Cggh90)
For sin(x)/1+cos(x), let sin(x) = f(x) and 1+ cos(x) = g(x)

f(x) tends to 0 as x tends to 0 and 1+cos(x) tends to 0 as x tends to 0.

So sin(x)/1+cos(x) tends to 0/2 = 0

Then the same can be done with |sinx/x| . |sinx/1+cos(x) finding that it tends to 0 ...

So 1-cos(x)/x tends to 0 ???
Yes. You are using various rules though (like that the quotient of two limits is equal to the limit of the quotient and that the product of two limits is equal to the limit of the products).

Incidentally, you are able to check this result is correct. If you recall from a-levels that and set and you can see intuitively what the limit should be. Note this is only intuition and not a rigorous proof because you are unlikely to have proven yet that the derivative of cosine is sine.

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