Find the limit if it exists:

http://rogercortesi.com/eqn/tempimagedir/eqn8721.png

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- March 3rd 2007, 11:22 PM^_^Engineer_Adam^_^Indeterminate Forms: Limits infinite... help
Find the limit if it exists:

http://rogercortesi.com/eqn/tempimagedir/eqn8721.png - March 3rd 2007, 11:58 PMJhevon
Now i dont know if i'm right, but maybe my answer can at least point you in the right direction.

Lim(x--> inf+) [sin(2/x)]/(1/x)

= Lim(x--> inf+) (2/2)[sin(2/x)]/(1/x)

= Lim(x--> inf+) 2[sin(2/x)]/(2/x)

Now as x--> infinity, 2/x -->0, so the above is equivalent to:

Lim(a-->0) 2sin(a)/a

= Lim(a-->0+) 2(sin(a)/a)

= 2*Lim(a-->0+) sin(a)/a

= 2*1 ..............................since lim(x-->0) sinx/x = 1

= 2

What bothers me is the whole approaching from the right thing, i dont know how that'll affect my solution - March 4th 2007, 12:31 AMticbol
Here is one way.

Lim(x--> +infinity)[sin(2/x) / (1/x)]

= sin(2/inf.) / (1/inf)

= sin(0) / (0)

= 0/0 ------------indeterminate.

Use the L'Hopital's Rule:

Lim(x-->a)[f(x) / g(x)] = Lim(x-->a)[f'(x) / g'(x)] for 0/0 limits.

So,

= Lim(x-->+inf.)[{cos(2/x) *(-2 /x^2)} / {-1 /x^2}]

= Lim(x-->+inf.)[cos(2/x) *2]

= 2cos(0)

= 2*1

= 2 ----------------answer. - March 4th 2007, 05:06 AMSoroban
Hello, ^_^Engineer_Adam^_^!

Quote:

Find the limit if it exists:

. . . . .sin(2/x)

. lim . --------

x→+∞ .(1/x)

Let u = 1/x

. . . . . . . . . . . . . . .sin(2u) . . . . . . . . .sin(2u)

And we have: . lim . -------- . = . lim . 2·--------- . = . 2(1) . = . 2

. . . . . . . . . . .u→0 . . .u . . . . .2u→0 . . . 2u