## yuki 5 years ago Find the limit as x -> infinity for the following

1. Yuki

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

2. Yuki

$\lim_{x \rightarrow \infty} x^{1/x}$

3. anonymous

first ones pinching theorm if I remember correctly lol

4. anonymous

and the second involves taking natural logs

5. 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

6. 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

7. 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

8. anonymous

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

9. anonymous

oh nvm , use elec's way ^_^

10. 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

11. anonymous

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

12. anonymous

Did you get the first one solved?

13. Yuki

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

14. anonymous

No, it's 1/2.

15. anonymous

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

16. anonymous

It's 0, not 2, though.

17. Yuki

I got the second one, that was perfect explanation.

18. anonymous

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

19. 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

20. anonymous

It does go to 0.

21. Yuki

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

22. Yuki

one of those typos maybe.

23. 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

24. anonymous

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

25. 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.

26. Yuki

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

27. Yuki

or was it e?

28. anonymous

similar to the second question I did

29. Yuki

I thought so

30. anonymous

and its just e

31. anonymous

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

32. Yuki

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

33. Yuki

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

34. anonymous

But e^x is still the same?

35. anonymous

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

36. Yuki

yep infinity to the zero power = 1 yaaay ! lol

37. anonymous

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

38. anonymous

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

39. anonymous

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

40. Yuki

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

41. anonymous

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

42. anonymous

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

43. anonymous

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

44. Yuki

you are such a math geek lol

45. anonymous

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

46. 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.

47. anonymous

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