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

**evankiefl** · March 9th 2011, 09:57 PM

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

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.

**FernandoRevilla** · March 9th 2011, 10:07 PM

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

**chisigma** · March 9th 2011, 10:49 PM

Originally Posted by evankiefl

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

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...

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

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

Kind regards

$\chi$ $\sigma$

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