yuki
  • yuki
Find the limit as x -> infinity for the following
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
yuki
  • yuki
\[\lim_{x \rightarrow \infty} \sin(x/2)/x\]
yuki
  • yuki
\[\lim_{x \rightarrow \infty} x^{1/x}\]
anonymous
  • anonymous
first ones pinching theorm if I remember correctly lol

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anonymous
  • anonymous
and the second involves taking natural logs
anonymous
  • anonymous
we know that -1<= sin(x/2) <= 1 divide through by x ( we can do this as x is not zero, it is approaching infinity!) so -1/x <= sin(x/2) / <= 1/x
anonymous
  • anonymous
now take the limits on the outside , as x-> infinity , they both go to zero , therefore the limit that is pinched between goes to zero
anonymous
  • anonymous
for second one, let the limit be L L = limit x->infinity ( x^(1/x) ) take ln of both sides so ln(L) = ln [ limit x->infinity x^(1/x) ] but the function "ln" is continuous as n->infinity so we can bring it inside the limit ( there is a continuity theorm that allows us to do this , I dont remember it right now
anonymous
  • anonymous
hmm, you can use L'hopital's rule for the first one since it's infinity/infinity
anonymous
  • anonymous
oh nvm , use elec's way ^_^
anonymous
  • anonymous
so ln(L) = limit (x->infinity ) ln[ x^(1/x) ] RHS = limit (x->infinity ) (1/x) ln(x) [ by a log law ] now this limit is an indeterminate form ( infinity /infinity ) so use La Hopitals rule RHS = limit (x->infinity ) (1/x) / 1 = limit (x->infinity ) (1/x) = 0 but RHS = ln(L) =0 so L= e^0 = 1 The limit is 1. Nice and long questions lol
anonymous
  • anonymous
the longer the question, the fun it is to solve =P
anonymous
  • anonymous
Did you get the first one solved?
yuki
  • yuki
I have bad news. the first one the answer is 2
anonymous
  • anonymous
No, it's 1/2.
anonymous
  • anonymous
Oh -- hang on...I thought you were taking it to 0. sorry!
anonymous
  • anonymous
It's 0, not 2, though.
yuki
  • yuki
I got the second one, that was perfect explanation.
anonymous
  • anonymous
This is a good summary of limit laws: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html
yuki
  • yuki
Let me double check the first one. I mean, intuitively sin(x/2) oscilates and x keeps growing so the limit seems to go to 0
anonymous
  • anonymous
It does go to 0.
yuki
  • yuki
you know what, I believe you guys. the answer said 1/2, but clearly x -> infinity should make it go to 0
yuki
  • yuki
one of those typos maybe.
anonymous
  • anonymous
you can sub in a really big value for x, say 1 000 000 and see that it works , it not a 100% mathematically correct proof by subing in numbers , but it should help you accept that the limit is zero
anonymous
  • anonymous
1 000 000 radians off course, need to be in correct mode , which you most likely wont be :P
anonymous
  • anonymous
You can say,\[\lim_{x \rightarrow \infty}\frac{\sin(x/2)}{x}=\lim_{x \rightarrow \infty} \sin(x/2).\lim_{x \rightarrow \infty}\frac{1}{x}=\lim_{x \rightarrow \infty} \sin(x/2).0=0\]since sin(x/2) is bounded for all x.
yuki
  • yuki
I would like to ask another limit question. how is it shown that \[\lim_{x \rightarrow \infty} (1+1/x)^x = e^x\]
yuki
  • yuki
or was it e?
anonymous
  • anonymous
similar to the second question I did
yuki
  • yuki
I thought so
anonymous
  • anonymous
and its just e
anonymous
  • anonymous
take logs , bring the power down , bring the limit inside the function, use LH's rule etc
yuki
  • yuki
would it be the same for \[\lim_{x \rightarrow \infty} (e^x +1/x )^x\]
yuki
  • yuki
why do I keep doing this? the last exponent is actually "1/x" not "x"
anonymous
  • anonymous
But e^x is still the same?
anonymous
  • anonymous
You'd attempt something similar. You have to remove the power.
yuki
  • yuki
yep infinity to the zero power = 1 yaaay ! lol
anonymous
  • anonymous
You can manipulate it into an indeterminate form and use L'Hopital's.
anonymous
  • anonymous
\[y=\left( e^x+\frac{1}{x} \right)^{1/x} \rightarrow \log y = \frac{1}{x}\log \left( e^x+\frac{1}{x} \right)=\frac{1/x}{1/\log \left( e^x+\frac{1}{x} \right)}\]
anonymous
  • anonymous
0/0 is the form as x goes to infinity.
yuki
  • yuki
I was able to do the lim = e thingy. I was always wondering how that works, thanks everyone.
anonymous
  • anonymous
Or...der...I could have left it as was noting it in the indeterminate form infty/infty.
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty} \log y =\lim_{x \rightarrow \infty} \frac{\log(e^x+1/x)}{x}=\frac{\lim_{x \rightarrow \infty} (\log (e^x+1/x))'}{\lim_{x \rightarrow \infty} (x)'}=\frac{\lim_{x \rightarrow \infty} \frac{e^x-\frac{1}{x^2}}{e^x+\frac{1}{x}}}{\lim_{x \rightarrow \infty} 1}\]
anonymous
  • anonymous
\[=\lim_{x \rightarrow \infty} \frac{e^x-\frac{1}{x^2}}{e^x+\frac{1}{x}}=\lim_{x \rightarrow \infty} \frac{1-\frac{1}{e^xx^2}}{1+\frac{1}{e^xx}}=1\]
yuki
  • yuki
you are such a math geek lol
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty}\log y = 1 \rightarrow y = e\]
yuki
  • yuki
Thanks everyone, it really helped me a lot. I'm going to take an AP calc diagnostic test soon and that will determine whether I can marry my GF or not. It's on Thursday and I think at this pace I can do a lot better than I thought. Again, thanks people.
anonymous
  • anonymous
You're welcome, Yuki! Good luck! :D

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