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

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

Mathematics
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I've tried several things, transformation formulae, series expansions etc etc. Very stuck.
\[\large \lim_{x\rightarrow 0 } \frac{\cos(\sin(x))-\cos(x)}{x^2}\]
On mobile, so thanks for the formatting.

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Split it into two limits, one should require squeeze theorem I believe
Squeeze theorem doesn't make sense if you have a x^2 term I thought?
No, it does.
we cannot split the limit here because the individual limits don't exist
have u tried taylor ? looks the numerator is \(\mathcal{O}(x^3)\)
I think you maybe right no we can just use L'hopital's rule
0/0
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 :)
looks good @qwerty!
:) thanks @Astrophysics
cos(A) \(\color{red}{\text{+}}\) cos(B)?
Yeah I did L'H twice and got 0 as well
Yup did same thing haha
Oh well, we're required to solve without L'Hospital. So anyone got any clues on that.
Any other restrictions we should know about...?
If you can use series expansions then saying you can't use L'H is practically meaningless imo
And I don't know what or how Taylor Series are... No other restrictions.
series expansion works nicely
just show that the degree of numerator is at least 3 and you're done
Niceee
Okay? @ganeshie8 and that's proof cause?
so are you allowed to use below ? \[\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots\] \[\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots\]
yes.
then use them
I've gone the next few steps... but I don't understand how that gives me a zero.!?!?
\[\large \cos(\sin x) = 1-\dfrac{(\sin x)^2}{2}+\cdots = 1 -\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\cdots \] fine with above ?
yes. got that step. Im afraid I need to be led by the hand still though.
\[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{1-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{x^2+\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{\mathcal{O}(x^4)}\right) \\~\\ &=\mathcal{O}(x^4) \end{align}}\]
that shows that the degree of each term in the numerator is at least \(4\)
I'm so sorry, but whats that o like symbol? starting out with calc
If it helps, you may replace \(\mathcal{O}(x^4)\) by \(x^4(stuff)\)
got it. thanks
this actually pretty clever.
\[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{1-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{x^2+x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{x^4(stuff)}\right) \\~\\ &=x^4(stuff) \end{align}}\]
oh no need for that. I worked it out too. thanks a lot lot lot
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I'm like that after every single calc question. so many...So many. I'm probably going to keep asking today
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Cute xD

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