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March 18th, 2018, 05:51 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 262 Thanks: 29  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 05:54 AM. 
March 18th, 2018, 11:27 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143 
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, 02:31 PM  #3  
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622  Quote:
 
March 18th, 2018, 02:59 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143  

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