Find the limit if it exists:
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
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.
Hello, ^_^Engineer_Adam^_^!
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