Question

find the limit as x→0 of cos(mx) − cos(nx) / x^2. use L'Hospital's rule if possible

Answer 1

\[\lim_{x \rightarrow 0}\frac{\cos(mx)-\cos(nx)}{x^2}\]

Answer 2

so can we use l'hospital?

Answer 3

we definitely know the bottom is 0 when plugin in 0

Answer 4

so only question to ask do we have 0/0

Answer 5

if so we can use l'hospital

Answer 6

I thought yes, because it's 1-1 for the numerator, which is 0.

Answer 7

yep yep

Answer 8

now we differentiate top
and differentiate bot

Answer 9

I know the bottom is 2x, but I'm confused about the top.

Answer 10

I found a solution online that pulls the m and the n out in front of their respective cosines, but I don't get why at all.

Answer 11

\[\lim_{x \rightarrow 0}\frac{\frac{d}{dx}(\cos(mx)-\cos(nx))}{\frac{d}{dx}(x^2)} \\ =\lim_{x \rightarrow 0}\frac{\frac{d}{dx} \cos(mx)-\frac{d}{dx}\cos(nx)}{2x}\]

Answer 12

well first of all you know the derivative of the outsides
that is what is the derivative of cos

Answer 13

Should I be treating cos(mx) and cos(nx) as a product rule?

Answer 14

well it isn't a product
it is a difference

Answer 15

\[\frac{d}{dx}(f-g)=\frac{d}{dx}f-\frac{d}{dx}g\]

Answer 16

so we use difference rule

Answer 17

but anyways to differentiate cos(mx)
we need to use chain rule

Answer 18

what is the derivative of cos

Answer 19

-sin is the derivative of cos

Answer 20

\[\lim_{x \rightarrow 0}\frac{\frac{d}{dx}(\cos(mx)-\cos(nx))}{\frac{d}{dx}(x^2)} \\ =\lim_{x \rightarrow 0}\frac{\frac{d}{dx} \cos(mx)-\frac{d}{dx}\cos(nx)}{2x} \\ =\lim_{x \rightarrow 0}\frac{(mx)'(-\sin(mx))-(nx)'(-\sin(nx))}{2x}\]
right and we also need to multiply by the derivative of the inside

Answer 21

\[(f(g(x))'=f'(x)|_{x=g(x)} \cdot g'(x) \\ (\cos(mx))'=(\cos(x))'|_{x=mx} \cdot (mx)'\]

Answer 22

\[(f(g(x))'=f'(x)|_{x=g(x)} \cdot g'(x) \\ (\cos(mx))'=(\cos(x))'|_{x=mx} \cdot (mx)' \\ (\cos(mx))'=-\sin(x)|_{x=mx} (m) \\ (\cos(mx))'=-\sin(mx) (m) \\ (\cos(mx))'=-m \sin(mx)\]

Answer 23

Okay, I understand that completely.

Answer 24

really ?
if so that is great

Answer 25

\[\lim_{x \rightarrow 0}\frac{\frac{d}{dx}(\cos(mx)-\cos(nx))}{\frac{d}{dx}(x^2)} \\ =\lim_{x \rightarrow 0}\frac{\frac{d}{dx} \cos(mx)-\frac{d}{dx}\cos(nx)}{2x} \\ =\lim_{x \rightarrow 0}\frac{(mx)'(-\sin(mx))-(nx)'(-\sin(nx))}{2x} \\ =\lim_{x \rightarrow 0}\frac{-m \sin(mx)+n \sin(nx)}{2x}\]
so what happens if we enter in 0 for x?

Answer 26

I believe you get -m * 0 + n * 0 = 0 for the numerator and 0 for the denominator, so you get 0/0 again. Which means you have to do L'Hospital's rule again

Answer 27

don't get frustrated
this will be the last round of l'hospital because the bottom will be 2

Answer 28

so we only have to differentiate that nasty top one more time

Answer 29

our new top anyways

Answer 30

we need to apply the chain rule again

Answer 31

hmmm, okay

Answer 32

do you want to try

Answer 33

just differentiate one term

Answer 34

I will try.

Answer 35

\[\frac{d}{dx}(-msin(mx))=\]

Answer 36

that -m in front is a constant multiply
you can drag it outside the derivative operator

Answer 37

\[\frac{d}{dx}\sin(mx) \text{ put main focus here }\]
we will multiply -m to the derivative later

Answer 38

So you'd get\[-m \cos(mx)*m\] ?

Answer 39

so it'd be \[-m ^{2}\cos(mx)\]

Answer 40

\[\frac{d}{dx}(-m \sin(mx)) \\ -m \frac{d}{dx}\sin(mx) \\ -m \cos(mx) \cdot m \\ -m^2 \cos(mx) \]
AWESOME! :)

Answer 41

\[\lim_{x \rightarrow 0}\frac{\frac{d}{dx}(\cos(mx)-\cos(nx))}{\frac{d}{dx}(x^2)} \\ =\lim_{x \rightarrow 0}\frac{\frac{d}{dx} \cos(mx)-\frac{d}{dx}\cos(nx)}{2x} \\ =\lim_{x \rightarrow 0}\frac{(mx)'(-\sin(mx))-(nx)'(-\sin(nx))}{2x} \\ =\lim_{x \rightarrow 0}\frac{-m \sin(mx)+n \sin(nx)}{2x} \\ \lim_{x \rightarrow 0} \frac{-m^2 \cos(mx)+n^2 \cos(nx)}{2}\]

Answer 42

you are ready to just plug in 0

Answer 43

YAY!!! So, cos(0) = 1, which means the final answer is \[\frac{ n ^{2}-m ^{2} }{ 2}\]

Answer 44

yah

Answer 45

Do you feel like anything here needs to be discussed more?

Answer 46

Thank you so much for your help! I'm set for now!

Answer 47

Ok cool
have a nice night

Answer 48

You too :)