Limit - which result is correct?

twoflower

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.

dextercioby

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??

twoflower

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

dextercioby

Can Maple compute limits??Can u compute derivatives??CAN U BOTH SOLVE THIS SIMPLE PROBLEM?????

Daniel.

ehild

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

twoflower

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.

HallsofIvy

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
tan³(x)- 3tan(x) is 3tan(x)sec²(x)- 3sec²(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(2²)/-1= -(36-12)= -24.

twoflower

I didn't want to prove the MAPLE calculation, I was trying to prove the teacher's calculation...

nrqed

That's indeed correct (notice that there's a typo, it should obviously be

"It's easy to calculate that the derivative of
tan³(x)- 3tan(x) is 3tan²(x)sec²(x)- 3sec²(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

ehild

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

twoflower

Nice ehild, I thought this is one of the limits which can't be solved without L'Hospital's rule...Thank you.