# Thread: L'Hopitals rule... getting right answer, but it should be negative not positive. Why?

1. ## L'Hopitals rule... getting right answer, but it should be negative not positive. Why?

Okay, so I'm working with L'Hopitals rule and indeterminate forms.. and I ran across this problem in my homework. I get the right answer, only that it is positive instead of negative and I can't figure out why it's negative.

The problem:
limit (sec(3x)cos(5x)) x-> (pi/2) from the left

since that is in infinity * 0 form... I did:

= cos(5x) / (1/sec(3x)) = 0/0

Great... I can now use L'Hopitals...

= -5sin(5x) / -3sin(3x) = 5/3

Negatives cancel right?? So now where did I go wrong? The answer should be negative according to my textbook, Wolfram Alpha, and MS Math.

2. Hello, nautica17!

There is a simple fact you overlooked . . . (I almost did!)

$\lim_{x\to\frac{\pi}{2}^-} \sec(3x)\cos(5x)$
Your work is correct . . .

Using L'Hopital, the problem becomes: . $\lim_{x\to\frac{\pi}{2}^-}\frac{5\sin(5x)}{3\sin(3x)}$

As $x$ approches $\tfrac{\pi}{2}$, the fraction approaches: . $\frac{5\sin\left(\frac{5\pi}{2}\right)}{3\sin\left (\frac{3\pi}{2}\right)} \;=\; \frac{5(1)}{3({\color{red}-1})} \;=\;-\frac{5}{3}$

3. Thank-you! That's so obvious now! I've been doing to many of these tonight. :|