Question

Is cosx a continuous function

Answer 1

Yes

Answer 2

No

Answer 3

its continuous , differentiable, second derivative exists, ... infinitely differentiable

Answer 4

\[\forall \epsilon \gt 0, \exists \delta \gt 0 : |x-a| \lt \delta \implies |\cos(x)-\cos(a)|\lt \epsilon\]

Answer 5

that is what you have to prove to show cos x is continuous

Answer 6

One can show that
\[
|\cos(a) - \cos(b)|\le |a-b|
\]
and that will do it

Answer 7

yeah i feel stuck after some point..

Answer 8

given an arbitrary epsilon, what will be your delta

Answer 9

ahh then its easy

Answer 10

\(\large |\cos x - \cos a| \le |x-a| \le \delta\)

so \(\large\delta = \epsilon \) ?

Answer 11

My method above has some cheating in it, because to prove the above inequality, you need the mean value theorem which requires that cos(x) is differentiable.
See this for a proof from the definition
http://www.math10.com/en/algebra/functions/continuity-sine-cosine-function/continuity-sin-cos-function.html

Answer 12

oh yeah we can't use MVT

Answer 13

The proof really uses the geometrical definition of cos(x) and sin(x)

Answer 14

MVT proves that inequality nicely though : \[\large \dfrac{\cos x - \cos a}{x-a} = (\cos (c))'\]

Answer 15

**right hand side should be f'(c)

Answer 16

Yes

Answer 17

In fact, the above inequality shows that cos(x) is uniformly continuous.

Answer 18

my notes has this for `uniform continuous` sir
\[\forall \epsilon \gt 0, \exists \delta \gt 0 , \forall a \in \mathbb{R}, \forall x\in \mathbb{R} : |x-a| \lt \delta \implies |\cos(x)-\cos(a)|\lt \epsilon \]

for continuous :
\[\forall a \in \mathbb{R}, \forall \epsilon \gt 0, \exists \delta \gt 0 , \forall x\in \mathbb{R} : |x-a| \lt \delta \implies |\cos(x)-\cos(a)|\lt \epsilon \]

Answer 19

im not too sure how changing the order of quantifiers changes the meaning