Thread: Does l'hospitals rule apply here?

I'm asked to evaluate the limit as x approaches pi/2 from the left of sec(7x)cos(3x). Would l'hospitals rule apply here or am I going about this wrong?

You can apply it anywhere but it is not necessary helpful.
$\displaystyle sec(7x)cos(3x) = \frac{cos(3x)}{cos(7x)}$

$\displaystyle lim_{x=\frac{\pi}{2}} \frac{cos(3x)}{cos(7x)} = lim_{x=\frac{\pi}{2}} \frac{-3sin(3x)}{-7sin(7x)} = \frac{-3sin(3\frac{\pi}{2})}{-7sin(7\frac{\pi}{2})} = \frac{3}{7}$

Well the reason I ask is that we just covered l'hospitals rule and this was on the homework we received, so I think perhaps we're supposed to use it if it applies

Well it applies that's what I showed you.
( diff(cos(u)) = -sin(u)du)

Originally Posted by vincisonfire

You can apply it anywhere but it is not necessary helpful.
$\displaystyle sec(7x)cos(3x) = \frac{cos(3x)}{cos(7x)}$

$\displaystyle lim_{x=\frac{\pi}{2}} \frac{cos(3x)}{cos(7x)} = lim_{x=\frac{\pi}{2}} \frac{-3sin(3x)}{-7sin(7x)} = \frac{-3sin(3\frac{\pi}{2})}{-7sin(7\frac{\pi}{2})} = \frac{3}{7}$

And how did you do that last step, simplifying to just 3/7 when the two terms following 3 and 7 in the num. and denom. are different?

Originally Posted by fattydq

And how did you do that last step, simplifying to just 3/7 when the two terms following 3 and 7 in the num. and denom. are different?

The value of $\displaystyle \sin{\frac{a\pi}{2}}$, assuming a is always a nonzero integer, will always equal 1, since sine is a periodic function. Go on, try it. So, he or she basically just canceled out 1 and 1 from the fraction.

Ahh thanks to both of you

Originally Posted by EightballLock

The value of $\displaystyle \sin{\frac{a\pi}{2}}$, assuming a is always a nonzero integer, will always equal 1, since sine is a periodic function. Go on, try it. So, he or she basically just canceled out 1 and 1 from the fraction.

Not quite... Depending on the remainder in the division of $\displaystyle a$ by 4, $\displaystyle \sin\frac{a\pi}{2}$ can take the values 0, 1 or -1. Go on, try it .
In the present case, the explanation is that $\displaystyle \frac{7\pi}{2}=\frac{3\pi}{2}+\frac{4\pi}{2}=\frac {3\pi}{2}+2\pi$ and $\displaystyle \sin(x+2\pi)=\sin x$ for any $\displaystyle x$ (i.e. sine is periodic, with period $\displaystyle 2\pi$). (By the way, $\displaystyle \sin\frac{3\pi}{2}=-1$).