Question

Help me please! Limit as x --> +∞ (sin(2x)-cos^2 (3x)) / x

Answer 1

Is that supposed to be
\[\lim_{x->\inf}\frac{\sin(2x)-\cos^2(3x)}{x}\]

Answer 2

If so, what is the maximum value the numerator can possibly take on?

Sorry, looks like that should be\[\lim_{x->\infty}\frac{\sin(2x)-\cos^2(3x)}{x}\]

Answer 3

I dont know

Answer 4

What is the range of the sin and cos functions?

Answer 5

Here's a hint:

Answer 6

Something to do with 1 and 0?

Answer 7

So if we found a value of \(x\) where \(\sin(2x) = 1\) and \(\cos^2(3x) = -1\), that would get us all the way up to \(2\) for the numerator. As it turns out, we don't even get that large, but the details are not important.

Our limit is effectively
\[\lim_{x->\infty}\frac{2}{x}\]
What's that limit?

Answer 8

So, they cant go beyond -1 and 1 so 1 and -1 are limits

Answer 9

What I am saying is that the numerator is going to be a number between -2 and 2, at the most. But we are dividing by an ever-increasing number as that limit goes to infinity. What is going to happen to the value of that fraction as we do so?

Answer 10

Umm... can you write me the answer and i cant fiure out the method to do it?

Answer 11

What is 1/10? What is 1/100? 1/1000? 1/10000? 1/10000000000? Are they approaching any particular value as the denominator gets larger and larger?

Answer 12

yes, 0

Answer 13

well, there you go. Your limit is effectively a small number divided by x as x goes to infinity. The behavior of the numerator doesn't ever do anything that could allow it to "keep up" with the ever-increasing denominator, so the limit is 0.

Answer 14

allright, so how do i use that to solve problems

Answer 15

Examine the behavior of numerator and denominator as the variable goes to the limit. Often you will be able to deduce the limit from this alone.

Answer 16

if denominator goes to infinity, unless the numerator also goes to infinity, your limit will be 0

if numerator goes to infinity and denominator does not, then you will not have a limit

if both go to infinity, other techniques will be needed

Answer 17

so, zeroes increase and number gets lower but it cant get lower than 0 and it approaches it infinetly

Answer 18

one definition of a limit is that (roughly speaking) for any amount of "closeness" you want, you can find a value of the variable that gets the value of the expression within that distance of the limit. No matter what tolerance value you give me, I can come up with a value of \(x\) that makes the expression closer to 0 than your tolerance value, so 0 is the limit.

Answer 19

and if limit is lets say 2, how do we know if it approaches it from 5 4 3 2.1 2.01 .... or from 1 1.9 1.99999

Answer 20

I learn almost anything by imagining it visually... i cant see math very well... Im an artist type

Answer 21

wouldn't you agree that?

\(\large\color{#000000}{\displaystyle-2<\sin(2x)-\cos^2x(3x)<2 }\)

Answer 22

\(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{\pm2}{x}=0 }\)

Answer 23

well how much is sin(2x)−cos2x(3x)

Answer 24

Perhaps my colleague, possessing as he does the wisdom of Solomon :-) can help you. I need to go...

Answer 25

therefore by the squeeze theorem

\(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~\infty} \frac{\sin(2x)-\cos^2x(3x)}{x} =0}\)

Answer 26

basically

\(\color{#000000}{\displaystyle -2< \sin(2x)-\cos^2x(3x)<2 }\)


\(\color{#000000}{\displaystyle\frac{-2}{x}<0 \frac{\sin(2x)-\cos^2x(3x)}{x} <\frac{2}{x} }\)

\(\color{#000000}{\displaystyle\lim_{x\to\infty}\frac{-2}{x}<\lim_{x \rightarrow ~\infty} \frac{\sin(2x)-\cos^2x(3x)}{x} <\lim_{x\to\infty}\frac{2}{x} }\)

Answer 27

And you can probably tell that

\(\color{#000000}{\displaystyle\lim_{x\to\infty}\frac{{\rm any ~~sines~~and~~cosines~~(sums/differences~of~them)} }{{\rm an{~}increasing{~}function}{~}f(x)}=0 }\)

Answer 28

even if this function is ln(x), (which is in fact the smallest -
via lim x-->∞ ln(x)/x^p=0; p>0, through L'H'S)

Answer 29

last reply can be disregarded.... good luck

Answer 30

Thanks...

Answer 31

Anytime!