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June 28th, 2017, 06:47 AM  #1 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
Hello everybody I appreciate your help with this problem: limit (((1+tan(x))^0.5(1+sin(x))^0.5)/x^3,x>0) I can solve this with L' Hôpital's Rule by differentiating the numerator and denominator 3 times and finding the limit of the resultant numerator and denominator when x>0 and dividing them: (3/2)/6=1/4! But I'd like to see whether there is another way of solving this problem without using L' Hôpital's Rule, like by some trick or linearization... it was assigned in part of the textbook that hasn't talked about L' Hôpital's Rule up to that point! Thank you for your help in advance... Last edited by skipjack; June 28th, 2017 at 07:26 AM. 
June 28th, 2017, 07:19 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,444 Thanks: 1462 
Hint: multiply numerator and denominator by cos(x)((1+tan(x))^0.5+(1+sin(x))^0.5).

June 28th, 2017, 07:29 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 192 Thanks: 23 
$\displaystyle x\rightarrow 0$ $\displaystyle \lim \frac{\sqrt{1+\tan{x}}\sqrt{1+\sin x}}{x^3}=\lim \frac{\tan{x}  \sin x}{2x^3}=\frac{1}{2} \lim \frac{\sin x(\frac{1}{\cos x}1)}{x^3}=$ $\displaystyle =\frac{1}{2} \lim \frac{(\frac{1}{\cos x}1)}{x^2}\cdot \lim \frac{\sin x}{x}=\frac{1}{2} \lim \frac{1\cos x}{x^2 \cos x}=\frac{1}{2} \lim \frac{1\cos x}{x^2}=\frac{1}{4} $ Last edited by skipjack; June 28th, 2017 at 09:25 AM. 
June 28th, 2017, 07:34 AM  #4 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
thanks, but I still don't know what to do with the x^3 that's left in the denominator..

June 28th, 2017, 07:37 AM  #5 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
I got it! thanks a lot idontknow and skipjack 
June 28th, 2017, 09:31 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,444 Thanks: 1462 
(1  cos(x))/x² = 2sin²(x/2)/(4(x/2)²) = (1/2)(sin(x/2)/(x/2))² 

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finding, hôpital, hôpitals, hopitals, limit, rule 
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