what is the lim of xsin(1/x) as x approaches infinity?

Question

Ok so here is what i did, i wrote it over as sin(1/x)/(1/x) the denominator is the reciprocal of x and if multiplied out will be the same. NOW does this equal one? That is the answer my teacher gave me (that it is one) but im not sure if that is the correct work. I graphed it on my calculator but im not getting a good picture. Is it supposed to be in radians?

What about if you have the same problem but have x^2 in the beginning instead of x? the answer my teacher gave me is infinity, but that would have to be since i do the same thing idid above.
limx^2 (sin1/x) as x approaches infinity.
(x sin(1/x)/(1/x)= x(sin(1/x)/(1/x))= x*1 --> infinity times 1 equals infinity. is this correct? for both problems?
THANKS!!!!! im really stressing out over this!

thanks,but we havnt learned l'hospital's rule yet, so i dnt understand, you just diferentiate both top and bottom?

Answer 1

Let me give you another perspective; I will use the series expansion definition of a sin(x)

We can define sin(x) as:

sin(x) = x - x^3/3! +x^5/5!- and so on...

Substitute 1/x for x in the expansion, you will get

sin(1/x) = 1/x - 1/(3!x^3) + 1/(5!x^5) -.....

multiply by x
xsin(1/x) = 1 - 1/(3!x^2) + 1/(5!x^4)-....

Let x go to infinite....the only term that survives is 1 because all of the other terms converge to zero.

So, your teacher was right,

The lim as x goes to infinite of xsin(1/x) = 1

If you got a x^2 in the expression, the result is infinite because the first factor in the series is 1/x that when multiplied by x^2 will leave an x...that will go to infinity as x goes to infinity.

Good Luck.

Answer 2

Hi kitty,

To find the lim of xsin(1/x) as x->infinity you can re-write it in the form that you did above, but when you look at the limits, you get 0/0
at this point, you can use l'hospital's rule which states that you can differentiate both the top and bottom function one again

so you will have
sin(1/x)/(1/x)

which will change to:
(-1/x^2*cos(1/x))/(-1/x^2)

-1/x^2 will cancel out and you are left with the
lim x-> infinity of cos(1/x) = 1(since the cos of 0 is 1)

---------

L'hospital's rule says that if you get anything in the form of 0/0 or infinity/infinity, you can differentiate anything that is above the divisor and below the divisor seperately.

Wikipedia has a good description of it here:
http://en.wikipedia.org/wiki/L%27hospital

Answer 3

The limit is 1. Use L'hopitals rule.

Answer 4

you need to apply l'hopitals rule you had the right approach:

rewrite it as lim sin(1/x) / (1/x)
now apply lhopital by taking the derivative of top and bottom
(-cos(1/x) / x²) / (-1/x²) = cos(1/x)

now lim cos(1/x) = cos (0) = 1

Answer 5

Hint: Use L'Hôpital's Rule.

Answer 6

Your first step was right. Then you can use l'Hopital's rule:
lim [f(x)/g(x)] = lim [f'(x)/g'(x)]. You'll get one as the answer.

Answer 7

x = a million ; (a million+a million/x)^(x) = (a million+a million/a million)^(a million) = 2 x = 10 ; (a million+x)^(a million/x) = (a million+a million/10)^(10) = 2.5937 x = a hundred ; (a million+x)^(a million/x) = (a million+a million/a hundred)^(a hundred) = 2.7408 x = one thousand ; (a million+x)^(a million/x) = (a million+a million/one thousand)^(one thousand) = 2.7169 x = ten thousand ; (a million+x)^(a million/x) = (a million+a million/ten thousand)^(ten thousand) = 2.7181 . . . we see that as x -> infinity, (a million+x)^(a million/x) extra appropriate approximates e

Answer 8

the answer is easily solved using calculus... maybe someone else knows exactly how