# Solving limits containing trig expressions

• Aug 29th 2012, 12:11 PM
PhizKid
Solving limits containing trig expressions
Q: $\lim_{x\rightarrow \frac{\pi}{2}} \frac{tan(2x)}{x - \frac{\pi}{2}}$

I was given a couple of hints: use substitution, and that there isn't any need for the tangent double angle formula.

I would have never thought to use substitution if I tried to solve this all day, and I had been trying to manipulate the tangent double angle formula for like an hour before I was told I didn't need it.

How do you know when to use methods like substitution and when not to use the double angle formulas? Especially when there are like 30 of these on an exam, I can't even solve this 1 problem within an hour and I have to solve 30 in 45 minutes.

Anyway, I got (actually I didn't, since I was told to do all of this and could never have though of any of this on my own):

$\lim_{x\rightarrow (x - h)} \frac{sin(2h + \pi)}{h \cdot cos(2h + \pi)}\\\lim_{x\rightarrow (x - h)} \frac{-sin(2h)}{-h \cdot cos(2h)}\\\lim_{x\rightarrow (x - h)} \frac{sin(2h)}{h \cdot cos(2h)}$

I don't know what I should do now. Should I convert back to tangent? Use double angle formulas? More substitution? Something else? How do you know what exactly to do at this point?
• Aug 29th 2012, 12:48 PM
MaxJasper
Re: Solving limits containing trig expressions
Take derivatives of numerator and denominator...limit is limit of this fraction which becomes $\lim_{x\to \frac{\pi }{2}} \, 2 \sec ^2(2 x)\to2$
• Aug 29th 2012, 12:50 PM
PhizKid
Re: Solving limits containing trig expressions
Derivatives are not allowed for this section in the course
• Aug 29th 2012, 12:58 PM
HallsofIvy
Re: Solving limits containing trig expressions
I would do this: let $u= x- \frac{\pi}{2}$. Then $x= u+ \frac{\pi}{2}$ and $2x= 2u+ \pi$. Now, tangent has period $\pi$ so that $tan(2u+ \pi)= tan(2u)$. With that substitution, the limit becomes $\lim_{u\to 0}\frac{tan(2u)}{u}= 2\lim_{u\to 0}\frac{sin(2u)}{2u}\frac{1}{cos(2u)}$. Now, make one more substitution- let y= 2u so that limit becomes $\lim_{y\to 0}\frac{sin(y)}{y}\frac{1}{cos(y)}$.

Can you finish that?
• Aug 29th 2012, 12:59 PM
Plato
Re: Solving limits containing trig expressions
Quote:

Originally Posted by PhizKid
Q: $\lim_{x\rightarrow \frac{\pi}{2}} \frac{tan(2x)}{x - \frac{\pi}{2}}$

$\frac{tan(2x)}{x - \frac{\pi}{2}}=\left(\frac{\sin(2x)}{x-\tfrac{\pi}{2}}\right)\sec(2x)$