Solving limits containing trig expressions

Q: $\displaystyle \lim_{x\rightarrow \frac{\pi}{2}} \frac{tan(2x)}{x - \frac{\pi}{2}}$

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):

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

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?

Re: Solving limits containing trig expressions

Take derivatives of numerator and denominator...limit is limit of this fraction which becomes $\displaystyle \lim_{x\to \frac{\pi }{2}} \, 2 \sec ^2(2 x)\to2$

Re: Solving limits containing trig expressions

Derivatives are not allowed for this section in the course

Re: Solving limits containing trig expressions

I would do this: let $\displaystyle u= x- \frac{\pi}{2}$. Then $\displaystyle x= u+ \frac{\pi}{2}$ and $\displaystyle 2x= 2u+ \pi$. Now, tangent has period $\displaystyle \pi$ so that $\displaystyle tan(2u+ \pi)= tan(2u)$. With that substitution, the limit becomes $\displaystyle \lim_{u\to 0}\frac{tan(2u)}{u}= 2\lim_{u\to 0}\frac{sin(2u)}{2u}\frac{1}{cos(2u)}$. Now, make one more substitution- let y= 2u so that limit becomes $\displaystyle \lim_{y\to 0}\frac{sin(y)}{y}\frac{1}{cos(y)}$.

Can you finish that?

Re: Solving limits containing trig expressions

Quote:

Originally Posted by

**PhizKid** Q: $\displaystyle \lim_{x\rightarrow \frac{\pi}{2}} \frac{tan(2x)}{x - \frac{\pi}{2}}$

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