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.
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'HÃ´spital's rule??
Can Maple compute limits??Can u compute derivatives??CAN U BOTH SOLVE THIS SIMPLE PROBLEM????? Daniel.
Your result is correct. 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
I don't understand. Using L'Hospital I got the result I wrote here. But the "official" result is different to mine. I was just asking how to prove it in Maple.
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= -(36-12)= -24.
That's indeed correct (notice that there's a typo, it should obviously be "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
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
Nice ehild, I thought this is one of the limits which can't be solved without L'Hospital's rule...Thank you.