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anonymous
 one year ago
limit x tends to 0 for f(x)= (cos(sinx)cosx)/x^2
anonymous
 one year ago
limit x tends to 0 for f(x)= (cos(sinx)cosx)/x^2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've tried several things, transformation formulae, series expansions etc etc. Very stuck.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \lim_{x\rightarrow 0 } \frac{\cos(\sin(x))\cos(x)}{x^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0On mobile, so thanks for the formatting.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Split it into two limits, one should require squeeze theorem I believe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Squeeze theorem doesn't make sense if you have a x^2 term I thought?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we cannot split the limit here because the individual limits don't exist

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1have u tried taylor ? looks the numerator is \(\mathcal{O}(x^3)\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2I think you maybe right no we can just use L'hopital's rule

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1what 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
 one year ago
Best ResponseYou've already chosen the best response.2looks good @qwerty!

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1:) thanks @Astrophysics

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0cos(A) \(\color{red}{\text{+}}\) cos(B)?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I did L'H twice and got 0 as well

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yup did same thing haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh well, we're required to solve without L'Hospital. So anyone got any clues on that.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Any other restrictions we should know about...?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0If you can use series expansions then saying you can't use L'H is practically meaningless imo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I don't know what or how Taylor Series are... No other restrictions.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1series expansion works nicely

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1just show that the degree of numerator is at least 3 and you're done

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay? @ganeshie8 and that's proof cause?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.0I've gone the next few steps... but I don't understand how that gives me a zero.!?!?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \cos(\sin x) = 1\dfrac{(\sin x)^2}{2}+\cdots = 1 \dfrac{(x  \frac{x^3}{3!}+\cdots)^2}{2}+\cdots \] fine with above ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes. got that step. Im afraid I need to be led by the hand still though.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\large{\begin{align} &\color{red}{ \cos(\sin x)}\color{blue}{\cos x}\\~\\ &= \color{red}{1\dfrac{(x  \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}\left(\color{blue}{1\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{\dfrac{(x  \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}\left(\color{blue}{\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{\dfrac{x^2+\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}\left(\color{blue}{\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{\dfrac{\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}\left(\color{blue}{\mathcal{O}(x^4)}\right) \\~\\ &=\mathcal{O}(x^4) \end{align}}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that shows that the degree of each term in the numerator is at least \(4\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm so sorry, but whats that o like symbol? starting out with calc

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1If it helps, you may replace \(\mathcal{O}(x^4)\) by \(x^4(stuff)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this actually pretty clever.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\large{\begin{align} &\color{red}{ \cos(\sin x)}\color{blue}{\cos x}\\~\\ &= \color{red}{1\dfrac{(x  \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}\left(\color{blue}{1\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{\dfrac{(x  \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}\left(\color{blue}{\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{\dfrac{x^2+x^4(stuff)}{2}+x^4(stuff)}\left(\color{blue}{\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{\dfrac{x^4(stuff)}{2}+x^4(stuff)}\left(\color{blue}{x^4(stuff)}\right) \\~\\ &=x^4(stuff) \end{align}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no need for that. I worked it out too. thanks a lot lot lot

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440488705192:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm like that after every single calc question. so many...So many. I'm probably going to keep asking today

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440489968673:dw
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