anonymous
  • anonymous
Help me please! Limit as x --> +∞ (sin(2x)-cos^2 (3x)) / x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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whpalmer4
  • whpalmer4
Is that supposed to be \[\lim_{x->\inf}\frac{\sin(2x)-\cos^2(3x)}{x}\]
whpalmer4
  • whpalmer4
If so, what is the maximum value the numerator can possibly take on? Sorry, looks like that should be\[\lim_{x->\infty}\frac{\sin(2x)-\cos^2(3x)}{x}\]
anonymous
  • anonymous
I dont know

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whpalmer4
  • whpalmer4
What is the range of the sin and cos functions?
whpalmer4
  • whpalmer4
Here's a hint:
1 Attachment
anonymous
  • anonymous
Something to do with 1 and 0?
whpalmer4
  • whpalmer4
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?
anonymous
  • anonymous
So, they cant go beyond -1 and 1 so 1 and -1 are limits
whpalmer4
  • whpalmer4
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?
anonymous
  • anonymous
Umm... can you write me the answer and i cant fiure out the method to do it?
whpalmer4
  • whpalmer4
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?
anonymous
  • anonymous
yes, 0
whpalmer4
  • whpalmer4
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.
anonymous
  • anonymous
allright, so how do i use that to solve problems
whpalmer4
  • whpalmer4
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.
whpalmer4
  • whpalmer4
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
anonymous
  • anonymous
so, zeroes increase and number gets lower but it cant get lower than 0 and it approaches it infinetly
whpalmer4
  • whpalmer4
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
I learn almost anything by imagining it visually... i cant see math very well... Im an artist type
SolomonZelman
  • SolomonZelman
wouldn't you agree that? \(\large\color{#000000}{\displaystyle-2<\sin(2x)-\cos^2x(3x)<2 }\)
SolomonZelman
  • SolomonZelman
\(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{\pm2}{x}=0 }\)
anonymous
  • anonymous
well how much is sin(2x)−cos2x(3x)
whpalmer4
  • whpalmer4
Perhaps my colleague, possessing as he does the wisdom of Solomon :-) can help you. I need to go...
SolomonZelman
  • SolomonZelman
therefore by the squeeze theorem \(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~\infty} \frac{\sin(2x)-\cos^2x(3x)}{x} =0}\)
SolomonZelman
  • SolomonZelman
basically \(\color{#000000}{\displaystyle -2< \sin(2x)-\cos^2x(3x)<2 }\) \(\color{#000000}{\displaystyle\frac{-2}{x}<0 \frac{\sin(2x)-\cos^2x(3x)}{x} <\frac{2}{x} }\) \(\color{#000000}{\displaystyle\lim_{x\to\infty}\frac{-2}{x}<\lim_{x \rightarrow ~\infty} \frac{\sin(2x)-\cos^2x(3x)}{x} <\lim_{x\to\infty}\frac{2}{x} }\)
SolomonZelman
  • SolomonZelman
And you can probably tell that \(\color{#000000}{\displaystyle\lim_{x\to\infty}\frac{{\rm any ~~sines~~and~~cosines~~(sums/differences~of~them)} }{{\rm an{~}increasing{~}function}{~}f(x)}=0 }\)
SolomonZelman
  • SolomonZelman
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)
SolomonZelman
  • SolomonZelman
last reply can be disregarded.... good luck
anonymous
  • anonymous
Thanks...
SolomonZelman
  • SolomonZelman
Anytime!

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