Limit of a Trig Function

• Jun 2nd 2010, 06:57 PM
EliteNewbz
Limit of a Trig Function
I'm quite confused on this question.

Limit Cos [ (pi * x) / (1- 4x) ]
X -> infinity

What I thought it was was to divide everything by x so it would be

Cos (pi / (1/x) - 4)

So as x -> inf 1/x = 0.

Cos (pi / 4)

am I doing this right?
• Jun 2nd 2010, 07:19 PM
Prove It
Quote:

Originally Posted by EliteNewbz
I'm quite confused on this question.

Limit Cos [ (pi * x) / (1- 4x) ]
X -> infinity

What I thought it was was to divide everything by x so it would be

Cos (pi / (1/x) - 4)

So as x -> inf 1/x = 0.

Cos (pi / 4)

am I doing this right?

Since the cosine function is continuous,

$\lim_{x \to \infty}\cos{\left(\frac{\pi x}{1 - 4x}\right)} = \cos{\left[\lim_{x \to \infty}\left(\frac{\pi x}{1 - 4x}\right)\right]}$

Since this goes to $\frac{\phantom{-}\infty}{-\infty}$ we can use L'Hospital's Rule, so

$\cos{\left[\lim_{x \to \infty}\left(\frac{\pi x}{1 - 4x}\right)\right]} = \cos{\left[\lim_{x \to \infty}\left(\frac{\phantom{-}\pi}{-4}\right)\right]}$

$= \cos{\left(-\frac{\pi}{4}\right)}$

$= \cos{\frac{\pi}{4}}$

$= \frac{\sqrt{2}}{2}$.
• Jun 3rd 2010, 12:42 AM
CaptainBlack
Quote:

Originally Posted by EliteNewbz
I'm quite confused on this question.

Limit Cos [ (pi * x) / (1- 4x) ]
X -> infinity

What I thought it was was to divide everything by x so it would be

Cos (pi / (1/x) - 4)

So as x -> inf 1/x = 0.

Cos (pi / 4)

am I doing this right?

Once you do what ProveIt does as a first step:

As $\cos$ is continuous:

$\lim_{x \to 0}\cos\left(\frac{\pi x}{1-4x}\right)=\cos\left[\lim_{x \to 0} \left(\frac{\pi x}{1-4x}\right)\right]$

yes.

CB