
#1
Jan1505, 12:07 PM

P: 368

Hi all,
I've been practising for the exam and did some limits from our master's collections: [tex] \lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan^{3} x  3\tan x}{\cos \left( x + \frac{\pi}{6} \right) } [/tex] I got the result 24, but there is written 12. Which one is correct? I tried to prove the result in Maple and Mathematica, but in neither I'm not able to get anything but symbolic result. Thank you. 



#2
Jan1505, 12:18 PM

Sci Advisor
HW Helper
P: 11,863

HINT:Bring your function to anither form,so that the limit can be put under the "conventional" form ">0".That requires a change of variable.
Daniel. P.S.Can as in do you know how to apply L'Hôspital's rule?? 



#3
Jan1505, 12:34 PM

P: 368

Ok, so now I put the "L'Hospitaled" function to the Maple and it still gives me symbolic result...




#4
Jan1505, 01:31 PM

Sci Advisor
HW Helper
P: 11,863

Limit  which result is correct?
Can Maple compute limits??Can u compute derivatives??CAN U BOTH SOLVE THIS SIMPLE PROBLEM?????
Daniel. 



#5
Jan1505, 03:29 PM

HW Helper
Thanks
P: 9,809

Just try to choose some value for x close enough to pi/3 and calculate the replacement value. You will see that it is far away from 12. ehild 



#6
Jan1505, 03:45 PM

P: 368





#7
Jan1505, 05:20 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

The point was: Why use MAPLE?? You can't PROVE the MAPLE calculation is correct by using MAPLE!
It's easy to calculate that the derivative of tan^{3}(x) 3tan(x) is 3tan(x)sec^{2}(x) 3sec^{2}(x) and certainly you should know that [itex]tan(\frac{\pi}{3})= \sqrt{3}[/itex] and that [itex]sec(\frac{\pi}{3}})= 2[/itex]. Of course, the derivative of [itex]cos(x+\frac{\pi}{6}[/itex] is [itex]sin(x+\frac{\pi}{6})[/itex] and [itex]sin(\frac{\pi}{2}})= 1[/itex]. The limit, by L'Hopital, is (3*3*4 3(2^{2})/1= (3612)= 24. 



#8
Jan1605, 01:39 AM

P: 368





#9
Jan1605, 12:41 PM

Sci Advisor
HW Helper
P: 2,886

"It's easy to calculate that the derivative of tan^{3}(x) 3tan(x) is 3tan^{2}(x)sec^{2}(x) 3sec^{2}(x)" i.e. there is a square in the first tan in the derivative but it's obviously a typo because HallsofIvy gave the correct numerical answer. Just a thought: if someone forgets to square the sec(x), they will get 12. So that could be the origin of the incorrect answer 12. Pat 



#10
Jan1605, 04:39 PM

HW Helper
Thanks
P: 9,809

The limit can be obtained even without L'Hopital's rule.
[tex] \lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan^{3} x  3\tan x}{\cos \left( x + \frac{\pi}{6} \right) }= \lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan(x)(\tan^{2} x  3)}{\cos ( x )\cos (\pi/6)\sin(x)\sin(\pi/6)}=[/tex] [tex]\lim_{x \rightarrow \frac{\pi}{3}}\frac{\tan(x)(tan(x)\sqrt(3))(\tan(x)+\sqrt(3))}{1/2\cos(x)(\sqrt(3)\tan(x))}= \lim_{x \rightarrow \frac{\pi}{3}}\frac{2\tan(x)(\tan(x)+\sqrt(3))}{\cos(x)}=\frac{2\sqrt(3)(2\sqrt(3)}{1/2}=24 [/tex] ehild 



#11
Jan1605, 04:59 PM

P: 368




Register to reply 
Related Discussions  
Limit problem  Correct?  Calculus & Beyond Homework  4  
Multivariable Limit, is my reasoning correct?  Calculus & Beyond Homework  2  
why is the limit 2025? a simple plug and chug limit! gone wrong!  Calculus & Beyond Homework  3  
time dilation(correct/not correct)  Introductory Physics Homework  1  
How to prove a limit is correct  Calculus  10 