Question

Find the limit as x -> infinity for the following

Answer 1

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

Answer 2

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

Answer 3

first ones pinching theorm if I remember correctly lol

Answer 4

and the second involves taking natural logs

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

oh nvm , use elec's way ^_^

Answer 10

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

Answer 11

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

Answer 12

Did you get the first one solved?

Answer 13

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

Answer 14

No, it's 1/2.

Answer 15

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

Answer 16

It's 0, not 2, though.

Answer 17

I got the second one, that was perfect explanation.

Answer 18

This is a good summary of limit laws:

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html

Answer 19

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

Answer 20

It does go to 0.

Answer 21

you know what, I believe you guys.

the answer said 1/2, but clearly x -> infinity
should make it go to 0

Answer 22

one of those typos maybe.

Answer 23

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

Answer 24

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

Answer 25

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.

Answer 26

I would like to ask another limit question.

how is it shown that \[\lim_{x \rightarrow \infty} (1+1/x)^x = e^x\]

Answer 27

or was it e?

Answer 28

similar to the second question I did

Answer 29

I thought so

Answer 30

and its just e

Answer 31

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

Answer 32

would it be the same for

\[\lim_{x \rightarrow \infty} (e^x +1/x )^x\]

Answer 33

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

Answer 34

But e^x is still the same?

Answer 35

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

Answer 36

yep

infinity to the zero power = 1 yaaay ! lol

Answer 37

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

Answer 38

\[y=\left( e^x+\frac{1}{x} \right)^{1/x} \rightarrow \log y = \frac{1}{x}\log \left( e^x+\frac{1}{x} \right)=\frac{1/x}{1/\log \left( e^x+\frac{1}{x} \right)}\]

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

\[\lim_{x \rightarrow \infty} \log y =\lim_{x \rightarrow \infty} \frac{\log(e^x+1/x)}{x}=\frac{\lim_{x \rightarrow \infty} (\log (e^x+1/x))'}{\lim_{x \rightarrow \infty} (x)'}=\frac{\lim_{x \rightarrow \infty} \frac{e^x-\frac{1}{x^2}}{e^x+\frac{1}{x}}}{\lim_{x \rightarrow \infty} 1}\]

Answer 43

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

Answer 44

you are such a math geek lol

Answer 45

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

Answer 46

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.

Answer 47

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