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March 18th, 2018, 04:51 AM   #1
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Basic limits

1)$\displaystyle \lim _{x\rightarrow 0} \frac{(x-\sin x )^n }{x^ 3} \; \; $, $\displaystyle n\in N$
2) $\displaystyle \lim _ {x\rightarrow 0} \frac{\cos x - \cos ^\sqrt{7} x}{x^2 }$

Last edited by idontknow; March 18th, 2018 at 04:54 AM.
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March 18th, 2018, 10:27 AM   #2
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You can use L'Hopital on (2) to get

$\lim \limits_{x\to 0} \dfrac{(x-\sin (x))^n}{x^3} = \dfrac{1}{2} \left(\sqrt{7}-1\right)$

(1) appears to be

$\lim \limits_{x\to 0} \dfrac{\cos (x)-\cos ^{\sqrt{7}}(x)}{x^2}= \begin{cases}\dfrac 1 6 &n=1 \\ \\ 0 &n>1 \end{cases}$
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March 18th, 2018, 01:31 PM   #3
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Quote:
Originally Posted by romsek View Post
You can use L'Hopital on (2) to get

$\lim \limits_{x\to 0} \dfrac{(x-\sin (x))^n}{x^3} = \dfrac{1}{2} \left(\sqrt{7}-1\right)$

(1) appears to be

$\lim \limits_{x\to 0} \dfrac{\cos (x)-\cos ^{\sqrt{7}}(x)}{x^2}= \begin{cases}\dfrac 1 6 &n=1 \\ \\ 0 &n>1 \end{cases}$
It looks like you switched your answers!
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March 18th, 2018, 01:59 PM   #4
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Quote:
Originally Posted by mathman View Post
It looks like you switched your answers!
you are correct sir!

flip those to be correct.
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