Limit - which result is correct?

[twoflower]

Hi all,

I've been practising for the exam and did some limits from our master's collections:

$$\lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan^{3} x - 3\tan x}{\cos \left( x + \frac{\pi}{6} \right) }$$

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]

I got the result -24, but there is written -12. Which one is correct?

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]

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

Daniel.

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 $tan(\frac{\pi}{3})= \sqrt{3}$ and that $sec(\frac{\pi}{3}})= 2$. Of course, the derivative of $cos(x+\frac{\pi}{6}$ is $-sin(x+\frac{\pi}{6})$ and $-sin(\frac{\pi}{2}})= -1$.

The limit, by L'Hopital, is (3*3*4- 3(2²)/-1= -(36-12)= -24.

 

[twoflower]

The point was: Why use MAPLE?? You can't PROVE the MAPLE calculation is correct by using MAPLE!

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

 

[nrqed]

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 $tan(\frac{\pi}{3})= \sqrt{3}$ and that $sec(\frac{\pi}{3}})= 2$. Of course, the derivative of $cos(x+\frac{\pi}{6}$ is $-sin(x+\frac{\pi}{6})$ and $-sin(\frac{\pi}{2}})= -1$.

The limit, by L'Hopital, is (3*3*4- 3(2²)/-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
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.

$$\lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan^{3} x - 3\tan x}{\cos \left( x + \frac{\pi}{6} \right) }= \lim_{x \rightarrow \frac{\pi}{3}} \frac{\tan(x)(\tan^{2} x - 3)}{\cos ( x )\cos (\pi/6)-\sin(x)\sin(\pi/6)}=$$

$$\lim_{x \rightarrow \frac{\pi}{3}}\frac{\tan(x)(tan(x)-\sqrt(3))(\tan(x)+\sqrt(3))}{1/2\cos(x)(\sqrt(3)-\tan(x))}= \lim_{x \rightarrow \frac{\pi}{3}}\frac{-2\tan(x)(\tan(x)+\sqrt(3))}{\cos(x)}=\frac{-2\sqrt(3)(2\sqrt(3)}{1/2}=-24$$

ehild

 

[twoflower]

The limit can be obtained even without L'Hopital's rule.
ehild

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