anonymous
  • anonymous
limit x tends to 0 for f(x)= (cos(sinx)-cosx)/x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I've tried several things, transformation formulae, series expansions etc etc. Very stuck.
Jhannybean
  • Jhannybean
\[\large \lim_{x\rightarrow 0 } \frac{\cos(\sin(x))-\cos(x)}{x^2}\]
anonymous
  • anonymous
On mobile, so thanks for the formatting.

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Astrophysics
  • Astrophysics
Split it into two limits, one should require squeeze theorem I believe
anonymous
  • anonymous
Squeeze theorem doesn't make sense if you have a x^2 term I thought?
Jhannybean
  • Jhannybean
No, it does.
ganeshie8
  • ganeshie8
we cannot split the limit here because the individual limits don't exist
ganeshie8
  • ganeshie8
have u tried taylor ? looks the numerator is \(\mathcal{O}(x^3)\)
Astrophysics
  • Astrophysics
I think you maybe right no we can just use L'hopital's rule
Astrophysics
  • Astrophysics
0/0
imqwerty
  • imqwerty
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 :)
Astrophysics
  • Astrophysics
looks good @qwerty!
imqwerty
  • imqwerty
:) thanks @Astrophysics
Jhannybean
  • Jhannybean
cos(A) \(\color{red}{\text{+}}\) cos(B)?
Empty
  • Empty
Yeah I did L'H twice and got 0 as well
Astrophysics
  • Astrophysics
Yup did same thing haha
anonymous
  • anonymous
Oh well, we're required to solve without L'Hospital. So anyone got any clues on that.
Empty
  • Empty
Any other restrictions we should know about...?
Empty
  • Empty
If you can use series expansions then saying you can't use L'H is practically meaningless imo
anonymous
  • anonymous
And I don't know what or how Taylor Series are... No other restrictions.
ganeshie8
  • ganeshie8
series expansion works nicely
ganeshie8
  • ganeshie8
just show that the degree of numerator is at least 3 and you're done
Astrophysics
  • Astrophysics
Niceee
anonymous
  • anonymous
Okay? @ganeshie8 and that's proof cause?
ganeshie8
  • ganeshie8
so are you allowed to use below ? \[\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots\] \[\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots\]
anonymous
  • anonymous
yes.
ganeshie8
  • ganeshie8
then use them
anonymous
  • anonymous
I've gone the next few steps... but I don't understand how that gives me a zero.!?!?
ganeshie8
  • ganeshie8
\[\large \cos(\sin x) = 1-\dfrac{(\sin x)^2}{2}+\cdots = 1 -\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\cdots \] fine with above ?
anonymous
  • anonymous
yes. got that step. Im afraid I need to be led by the hand still though.
ganeshie8
  • ganeshie8
\[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{1-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{x^2+\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{\mathcal{O}(x^4)}\right) \\~\\ &=\mathcal{O}(x^4) \end{align}}\]
ganeshie8
  • ganeshie8
that shows that the degree of each term in the numerator is at least \(4\)
anonymous
  • anonymous
I'm so sorry, but whats that o like symbol? starting out with calc
ganeshie8
  • ganeshie8
If it helps, you may replace \(\mathcal{O}(x^4)\) by \(x^4(stuff)\)
anonymous
  • anonymous
got it. thanks
anonymous
  • anonymous
this actually pretty clever.
ganeshie8
  • ganeshie8
\[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{1-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{x^2+x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{x^4(stuff)}\right) \\~\\ &=x^4(stuff) \end{align}}\]
anonymous
  • anonymous
oh no need for that. I worked it out too. thanks a lot lot lot
Astrophysics
  • Astrophysics
|dw:1440488705192:dw|
anonymous
  • anonymous
I'm like that after every single calc question. so many...So many. I'm probably going to keep asking today
ganeshie8
  • ganeshie8
|dw:1440489968673:dw|
Astrophysics
  • Astrophysics
Cute xD

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