# Thread: Calculation of a limit

1. ## Calculation of a limit

Hey

I'm not managing to calculate the following limit:

The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)

L'hopital's rule doesn't work because the limit of the quotient of the derivatives doesn't exist, and I haven't managed to find any trig identities that work. I understand intuitively that the limits is 1, since cosx and sinx are bounded and x isn't. But how can I get the function into a form where it's obvious?

2. Originally Posted by moses
Hey

I'm not managing to calculate the following limit:

The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)

L'hopital's rule doesn't work because the limit of the quotient of the derivatives doesn't exist, and I haven't managed to find any trig identities that work...
Hint: Divide numerator and denominator by x.

3. Originally Posted by moses
Hey

I'm not managing to calculate the following limit:

The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)

L'hopital's rule doesn't work because the limit of the quotient of the derivatives doesn't exist, and I haven't managed to find any trig identities that work. I understand intuitively that the limits is 1, since cosx and sinx are bounded and x isn't. But how can I get the function into a form where it's obvious?
Your 'intuition' is good because is...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. L'Hopital's rule does not apply because the limit is not of the form 0/0 or the variations on that. As x goes to 0, x+ cos(x)/2 goes to 1/2, not 0. Since the denominator does go to 0, the limit of the fraction does not exist.

5. Originally Posted by moses
The limit as x goes to infinity of (x + cos(x)/2) / (x - sin(x)/2)
Originally Posted by HallsofIvy
L'Hopital's rule does not apply because the limit is not of the form 0/0 or the variations on that. As x goes to 0, x+ cos(x)/2 goes to 1/2, not 0. Since the denominator does go to 0, the limit of the fraction does not exist.
The limit in the OP is as x goes to infinity, not 0.

-Dan

6. Okay thanks