Question

\[\lim_{x \rightarrow 0} \frac{ \cos(sinx)-\cos x }{ x^4 }\]

Answer 1

HI CODY!

Answer 2

hello

Answer 3

@ganeshie8 ??

Answer 4

@ParthKohli ?? @Nurali ??

Answer 5

@mathmale ?

Answer 6

have you gone over l'hospitals rule? @cody_123

Answer 7

i think that would be tedious

Answer 8

it sure would. but can you think of a better way of doing it at the moment ?
I am all ears

Answer 9

actually it wouldn't be too bad. you would have to differentiate the top 4 times.
if you're good with chain rule, it wouldn't be too bad.

Answer 10

\[\lim_{x \rightarrow 0}\frac{ -\sin(sinx)*\cos x+\sin x }{ 4x^3 }\]

Answer 11

good that's right

Answer 12

you will continue doing it until there is no more x in the denom

Answer 13

that wud be too long

Answer 14

alrighty... let's see if we can find another way to do it.

Answer 15

I'm looking into the special sine and cosine limits

Answer 16

cosC - cosD is not simplifying much :/

Answer 17

it sure isn't. @ganeshie8
@cody_123 I'm still working on it. Even if we can't find a better way of doing it, just know that you will come out with the correct solution using the rule.

Answer 18

yeah, it's crummy, but really it's gonna be l'hosital's rule.
I hate to be the bearer of bad news >.<
and it gets ugly

Answer 19

Continuous use of L'H Rule will be best.

Answer 20

it would be too long, any short method

Answer 21

Just a thought:

You can replace \(x^4\) with \(\sin^4(x)\) since \(\lim_{x \to 0}\frac{\sin(x)}{x} = 1\) and all.

But I don't really see how that would help either. Maybe we could use some more results like the sin(x)/x one.

Answer 22

I think it's 1/6. what I ended up doing was taking really small numbers close to 0 and estimating it in my calc
and it looks like it'll be 1/6.

Answer 23

http://www.wolframalpha.com/input/?i=lim_%28x+tends+to+0%29+%28cos%28sinx%29+-+cosx%29%2Fx%5E4+

Answer 24

1/6 .. is the answer lol

Answer 25

We need to get to 1/6 somehow

Answer 26

what about using expansions?

Answer 27

using tables on my calc it looks even more clear to me 1/6 should be it

Answer 28

like \[\cos x=1-\frac{ x^2 }{ 2! }+\frac{ x^4 }{ 4! }-....\]

Answer 29

oh, like a power series.

Answer 30

if we can expand cos(sin x) till we get x^4 we can cancel x^4 in numerator and denominator

Answer 31

but wouldn't that also get extremely convoluted?

Answer 32

Why not take y = sin(x) ..

Answer 33

expand until we get x^4
but that would be using an approximation since it's an infinite series

Answer 34

I like how you're thinking outside the box : )

Answer 35

but I don't think that approach would turn into anything we could use.

Answer 36

\[\dfrac{\cos(\sin(x))}{\sin^4 (x)} - \dfrac{\cot(x)}{\sin^3 (x)}\]HALP ME. PLS.

Answer 37

@mathslover oh okay, so using a substitution.
but then x would be sin^-1(y)

Answer 38

haha! I'm telling you, it's a really ugly problem and lhospital's looks like the only method that will reach an end

Answer 39

Well if I use substitution then the denominator will become :

\((\sin^{-1} y)^4\)

Differentiate it now... (w.r.t y)

\(\cfrac{4 \sin^{-1} (y)^3 }{\sqrt{1-y^2}} \)

Answer 40

Seems like this is getting uglier and uglier

ahah @Miracrown yeah..!

Answer 41

I honestly would discuss this one with your instructor and see if there are any tricks other than estimation to make the process go faster than lhospitals
other than estimation I'm just not seeing another way.

Answer 42

\(\lim_{y \rightarrow 0} \cfrac{-(\sin y) \sqrt{1-y^2} + y}{4\sin^{-1} (y^3) } \)

Answer 43

In this problem I'm not seeing a faster way. if you could use a calculator it would be pretty quick to use a table but I totally get that you probably can't.

Answer 44

oh wait I might be onto something
now it's my turn to try something!!

Answer 45

\[\cos x=1-\frac{ x^2 }{ 2! }+\frac{ x^4 }{ 4! }-..\]\[\sin x=x-\frac{ x^3 }{ 3! }+...\]

Answer 46

using the definition of derivative may yueld something

Answer 47

Expansion series may work now.

Answer 48

I'm trying to use the special limit
lim y-->0 [cos(y) - 1]/y = 0