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

\lim_{x \to 0} f(x) = f(0)=2

In short, yes, and the reason is continuity.

Reply 7

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?!

Reply 8

kfkle

Original post by Cggh90

Have a look at theorem 2.6 in your notes.

Reply 9

Cggh90

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!

Reply 10

kfkle

Original post by Cggh90

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?

Reply 11

Cggh90

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

(edited 13 years ago)

Reply 12

IrrationalNumber

Which university are you both at?

Reply 13

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 ???

Reply 14

kfkle

Original post by Cggh90

good job on the tests. averaging just over 90% on the assignments atm, how about you?

@IrrationalNumber: southampton university.

Reply 15

Cggh90

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?

Reply 16

Cggh90

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 :biggrin:

Reply 17

kfkle

Original post by Cggh90

I do happen to have 5 A's yes, but that's not why I'm here. it's a long story.

Reply 18

IrrationalNumber

Original post by Cggh90

We don't actually know each other IRL, but Southampton university..

Where did yo/do you study?

Warwick. Still there.

Reply 19

IrrationalNumber

Original post by Cggh90

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

f'(y)=\lim_{x \to y} \frac{f(x)-f(y)}{x-y}

and set

f=\cos

and

y=0

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.