Find the limit as x -> infinity for the following

- yuki

Find the limit as x -> infinity for the following

- katieb

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- yuki

\[\lim_{x \rightarrow \infty} \sin(x/2)/x\]

- yuki

\[\lim_{x \rightarrow \infty} x^{1/x}\]

- anonymous

first ones pinching theorm if I remember correctly lol

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## More answers

- anonymous

and the second involves taking natural logs

- 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

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

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

hmm, you can use L'hopital's rule for the first one since it's infinity/infinity

- anonymous

oh nvm , use elec's way ^_^

- 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

the longer the question, the fun it is to solve =P

- anonymous

Did you get the first one solved?

- yuki

I have bad news.
the first one the answer is 2

- anonymous

No, it's 1/2.

- anonymous

Oh -- hang on...I thought you were taking it to 0. sorry!

- anonymous

It's 0, not 2, though.

- yuki

I got the second one, that was perfect explanation.

- anonymous

This is a good summary of limit laws:
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html

- 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

It does go to 0.

- yuki

you know what, I believe you guys.
the answer said 1/2, but clearly x -> infinity
should make it go to 0

- yuki

one of those typos maybe.

- 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

1 000 000 radians off course, need to be in correct mode , which you most likely wont be :P

- 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

I would like to ask another limit question.
how is it shown that \[\lim_{x \rightarrow \infty} (1+1/x)^x = e^x\]

- yuki

or was it e?

- anonymous

similar to the second question I did

- yuki

I thought so

- anonymous

and its just e

- anonymous

take logs , bring the power down , bring the limit inside the function, use LH's rule etc

- yuki

would it be the same for
\[\lim_{x \rightarrow \infty} (e^x +1/x )^x\]

- yuki

why do I keep doing this?
the last exponent is actually "1/x" not "x"

- anonymous

But e^x is still the same?

- anonymous

You'd attempt something similar. You have to remove the power.

- yuki

yep
infinity to the zero power = 1 yaaay ! lol

- anonymous

You can manipulate it into an indeterminate form and use L'Hopital's.

- 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

0/0 is the form as x goes to infinity.

- yuki

I was able to do the lim = e thingy.
I was always wondering how that works, thanks everyone.

- anonymous

Or...der...I could have left it as was noting it in the indeterminate form infty/infty.

- 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

\[=\lim_{x \rightarrow \infty} \frac{e^x-\frac{1}{x^2}}{e^x+\frac{1}{x}}=\lim_{x \rightarrow \infty} \frac{1-\frac{1}{e^xx^2}}{1+\frac{1}{e^xx}}=1\]

- yuki

you are such a math geek lol

- anonymous

\[\lim_{x \rightarrow \infty}\log y = 1 \rightarrow y = e\]

- 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

You're welcome, Yuki! Good luck! :D

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