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 March 18th, 2018, 04:51 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 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.
 March 18th, 2018, 10:27 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,090 Thanks: 1086 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 $\lim \limits_{x\to 0} \dfrac{\cos (x)-\cos ^{\sqrt{7}}(x)}{x^2}= \begin{cases}\dfrac 1 6 &n=1 \\ \\ 0 &n>1 \end{cases}$
March 18th, 2018, 01:31 PM   #3
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 Originally Posted by romsek 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 $\lim \limits_{x\to 0} \dfrac{\cos (x)-\cos ^{\sqrt{7}}(x)}{x^2}= \begin{cases}\dfrac 1 6 &n=1 \\ \\ 0 &n>1 \end{cases}$

March 18th, 2018, 01:59 PM   #4
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you are correct sir!

flip those to be correct.

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