# Thread: Calculus 1: Limit of a function with sinx and cosx involved

1. ## Calculus 1: Limit of a function with sinx and cosx involved

$\displaystyle limit x---> infinity (x+sinx)/(3x+cosx)$

The oscillating nature of sin and cos lead me to believe that sinx and cosx don't have limits at infinity. But if not, I don't know what to do. I want to try using L'hopitals but as far I can tell this isn't in an indeterminate form.

2. Although $\displaystyle \lim_{x\to +\infty}\sin x$ does not exist we have $\displaystyle |\sin x|\leq 1$ for all $\displaystyle x\in \mathbb{R}$ . This means that $\displaystyle \lim_{x\to +\infty}(x+\sin x)=+\infty$ . Same considerations for the denominator.

3. Originally Posted by evankiefl
$\displaystyle limit x---> infinity (x+sinx)/(3x+cosx)$

The oscillating nature of sin and cos lead me to believe that sinx and cosx don't have limits at infinity. But if not, I don't know what to do. I want to try using L'hopitals but as far I can tell this isn't in an indeterminate form.
What You have to do is simply to devide numerator and denominator by x obtaining...

$\displaystyle \displaystyle \frac{x+ \sin x}{3x + \cos x} = \frac{1 + \frac{\sin x}{x}}{3+\frac{\cos x}{x}}$

... and then find the limit if x tends to infinity...

Kind regards

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