Question

limit x tends to 0 for f(x)= (cos(sinx)-cosx)/x^2

Answer 1

I've tried several things, transformation formulae, series expansions etc etc. Very stuck.

Answer 2

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

Answer 3

On mobile, so thanks for the formatting.

Answer 4

Split it into two limits, one should require squeeze theorem I believe

Answer 5

Squeeze theorem doesn't make sense if you have a x^2 term I thought?

Answer 6

No, it does.

Answer 7

we cannot split the limit here because the individual limits don't exist

Answer 8

have u tried taylor ?
looks the numerator is \(\mathcal{O}(x^3)\)

Answer 9

I think you maybe right no we can just use L'hopital's rule

Answer 10

0/0

Answer 11

what i did -

1) simplified cosA+cosB
2)took the constants out ...... only constant which ws there ws -2 :P
3)applied L hospitality
nd then i got (-2)x0
=0

:)

Answer 12

looks good @qwerty!

Answer 13

:) thanks @Astrophysics

Answer 14

cos(A) \(\color{red}{\text{+}}\) cos(B)?

Answer 15

Yeah I did L'H twice and got 0 as well

Answer 16

Yup did same thing haha

Answer 17

Oh well, we're required to solve without L'Hospital. So anyone got any clues on that.

Answer 18

Any other restrictions we should know about...?

Answer 19

If you can use series expansions then saying you can't use L'H is practically meaningless imo

Answer 20

And I don't know what or how Taylor Series are... No other restrictions.

Answer 21

series expansion works nicely

Answer 22

just show that the degree of numerator is at least 3 and you're done

Answer 23

Niceee

Answer 24

Okay? @ganeshie8 and that's proof cause?

Answer 25

so are you allowed to use below ?

\[\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots\]
\[\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots\]

Answer 26

yes.

Answer 27

then use them

Answer 28

I've gone the next few steps... but I don't understand how that gives me a zero.!?!?

Answer 29

\[\large \cos(\sin x) = 1-\dfrac{(\sin x)^2}{2}+\cdots = 1 -\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\cdots \]

fine with above ?

Answer 30

yes. got that step. Im afraid I need to be led by the hand still though.

Answer 31

\[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{1-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\
&= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\
&= \color{red}{-\dfrac{x^2+\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\
&= \color{red}{-\dfrac{\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{\mathcal{O}(x^4)}\right) \\~\\
&=\mathcal{O}(x^4)
\end{align}}\]

Answer 32

that shows that the degree of each term in the numerator is at least \(4\)

Answer 33

I'm so sorry, but whats that o like symbol? starting out with calc

Answer 34

If it helps, you may replace \(\mathcal{O}(x^4)\) by \(x^4(stuff)\)

Answer 35

got it. thanks

Answer 36

this actually pretty clever.

Answer 37

\[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{1-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\
&= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\
&= \color{red}{-\dfrac{x^2+x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\
&= \color{red}{-\dfrac{x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{x^4(stuff)}\right) \\~\\
&=x^4(stuff)
\end{align}}\]

Answer 38

oh no need for that. I worked it out too. thanks a lot lot lot

Answer 39

I'm like that after every single calc question. so many...So many. I'm probably going to keep asking today