Question

Limit[(Cos[t + h] - Cos[t])/h, h -> 0]

Answer 1

one way to do this is first use the identify

cos( a+b) = cos(a) cos(b) - sin(a) sin(b)

to write

cos(t+h) = cos(t) cos(h) - sin(t) sin(h)

next, we can use the series definition of cos and sin
see https://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions

in particular, write
cos(h) = 1 - h^2/2 ..... and higher order terms
sin(h)= h - h^3/6 .... and higher order terms

use those definitions in
\[ \frac{\cos(t) \cos(h) - \sin(t) \sin(h) - \cos(t)}{h} \]
you should be able to divide h evenly into the numerator
then take the limit as h->0

Answer 2

replace cos(h) with ( 1 - h^2/2 +....)
and sin(h) with ( h - h^3/6 ....)

Answer 3

hmm

Answer 4

how about
\[\lim_{h \rightarrow 0}\frac{ \cos h - 1 }{ h }=0\text{, and }\lim_{h \rightarrow 0}\frac{ \sin h }{ h }=1\]
that is hopefully something they've seen previously

Answer 5

so the answer will be -sin(t)

Answer 6

yep

Answer 7

pgpilot326 I sorta went that route and... thus far have...
\(\bf lim_{h \to 0}\cfrac{cos(t+h)-cos(t)}{h}\\ \quad \\ \quad \\
\cfrac{cos(t+h)-cos(t)}{h}\implies \cfrac{cos(t)cos(h)-sin(t)sin(h)-cos(t)}{h}\\ \quad \\
\textit{getting common factor atop}\\ \quad \\
\cfrac{cos(t)[cos(h)-1]-sin(t)sin(h)}{h}\\ \quad \\
\cfrac{cos(t)[cos(h)-1]}{h} - \cfrac{sin(t)sin(h)}{h}\\ \quad \\
\cfrac{cos(t)}{h}\cdot \cfrac{cos(h)-1}{h}- \cfrac{sin(t)}{h}\cdot \cfrac{sin(h)}{h} \implies -\cfrac{sin(t)}{h}\)

Answer 8

no that last line is incorrect. shouldn't have h twice in the denominator

Answer 9

ohh yeah!

Answer 10

oh using l hopital's rule right?

Answer 11

man... yes indeed

Answer 12

sweet thanks guys... =)

Answer 13

\(\bf lim_{h \to 0}\cfrac{cos(t+h)-cos(t)}{h}\\ \quad \\ \quad \\
\cfrac{cos(t+h)-cos(t)}{h}\implies \cfrac{cos(t)cos(h)-sin(t)sin(h)-cos(t)}{h}\\ \quad \\
\textit{getting common factor atop}\\ \quad \\
\cfrac{cos(t)[cos(h)-1]-sin(t)sin(h)}{h}\\ \quad \\
\cfrac{cos(t)[cos(h)-1]}{h} - \cfrac{sin(t)sin(h)}{h}\\ \quad \\
\cfrac{cos(t)}{h}\cdot \cfrac{cos(h)-1}{h}- sin(t) \cdot \cfrac{sin(h)}{h} \implies -sin(t)\)

Answer 14

1st term too

Answer 15

hehe, you're right...

Answer 16

\(\bf cos(t)\cdot \cfrac{cos(h)-1}{h}- sin(t) \cdot \cfrac{sin(h)}{h} \implies -sin(t)\)

Answer 17

that's it, sans the limit

Answer 18

Limit[(Cos[t + h] - Cos[t])/h, h -> 0]

is going back to the definition of a derivative.

\[ \frac{ \cos(t) \cos(h) - \sin(t) \sin(h) - \cos(t)}{h} \\
\frac{ \cos(t)( 1 - \frac{h^2}{2}+...) - \sin(t)(h - \frac{h^3}{3!}+...) - \cos(t)}{h} \\
\frac{ \cancel{\cos(t)} - \frac{h^2}{2} \cos(t)+... - h\sin(t) - \frac{h^3}{3!}\sin(t)+... - \cancel{\cos(t)} }{h} \\
- \frac{h}{2}\cos(t)+... - \sin(t) - \frac{h^2}{6} \sin(t)...
\]
now take the limit as h->0 and all the terms drop out except -sin(t)

Answer 19

yes but i3lue probably hasn't seen series yet. how does i3lue know that the series representations of sin and cos are valid?

Answer 20

are those factorial?

Answer 21

yes

Answer 22

i see what pi is trying to say .... hahaha its been awhile since i did series lol =P

Answer 23

yeah, what he's saying is perfectly valid

Answer 24

Limit[(Cos[t + h] - Cos[t])/h, h -> 0]
why not apply l'hopitals rule from the start, if you're going to use it later?

differentiate w.r.t. h...

Answer 25

That might not be perfectly valid, but it works... treating t as a constant.
Limit[(Cos[t + h] - Cos[t])/h, h -> 0]

becomes

Limit[-sin(t+h))/1, h -> 0]

Answer 26

Now plug in h=0, and you get -sin(t)

Answer 27

I mean, t is a constant... it's h that's approaching zero, t is constant.

Answer 28

Thanks =)

Answer 29

I mean i guess my way kinda circumvents the actual definition of the derivative, since you need to use actual derivative rules...