# Calculation of a limit

• April 28th 2011, 02:22 AM
moses
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?
• April 28th 2011, 02:29 AM
alexmahone
Quote:

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.
• April 28th 2011, 02:35 AM
chisigma
Quote:

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...

http://quicklatex.com/cache3/ql_ac1c...377eed7_l3.png

Kind regards

$\chi$ $\sigma$
• April 28th 2011, 03:58 AM
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.
• April 28th 2011, 04:17 AM
topsquark
Quote:

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

Quote:

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
• April 28th 2011, 05:21 AM
moses
Okay thanks :)