What is the lim of x^2 sin(1/x) as x approaches infinity?

Question

Im stuck! I found similar examples online but I dont understand them...any help is greatly appreciated!!! :) thnx

would it have the same answer of 0 if it was just x instead of x^2?

Answer 1

Hmmm, let's see....



Rewrite it as x times sin (1/x) / (1/x). I did that because I know something about the limit of (sin y)/y: as y approaches zero, the limit approaches 1.

The expression sin (1/x) / (1/x) is similar.

Ok, as x approaches infinity, then 1/x approaches 0. So let y = 1/x. Then that part of the problem becomes the limit of sin y / y as y approaches zero (when x aproaches infinity).

In other words,







lim ( x * sin(1/x) / (1/x) )
x -> infinity

equals

lim (x)
x -> inf

times

lim(sin (1/x) / (1/x))
x -> inf



The first limit is unbounded (obviously). The second limit is 1, because with the substitution y = 1/x, that limit becomes

lim (sin y / y)
y -> 0

which equals 1.

So you get infinity times 1, which is infinity.



Did that make any sense?

EDIT: The above arguments have two flaws. the first is assuming infinity times zero is zero. Not always. Sometimes, but not always.

The second is assuming that the exponent in your function is

2sin(1/x)

when i BET you function is really (x^2) * sin(1/x)

Answer 2

As x approaches infinity 1/x will tend toward zero and sin(0) = 0 and x^0 = 1 so the answer is one.

Answer 3

Graphically speaking, y = sin(u) looks like the linear y = u in the neighborhood of u = 0. So we can use the substitution u = 1/x to see that y = sin(1/x) looks like the linear y = 1/x as x tends toward infinity. Let's treat it as such.
The limit, then, of (1/x)(x^2) as x goes to infinity reduces to the limit of x as x goes to infinity, or an infinite limit.

Answer 4

let u = 1/x
then x = 1/u
as x approaches infinity, u approaches 0

using this info you can rewrite the original problem as:
The limit as u approaches 0 of sin(u)/u^2

This is an indeterminant form , lim = 0/0, so you can use L'Hopital's rule.

cos(u)/ 2u

As u approaches 0, cos(u) approaches 1 and 2u approaches zero. So the limit is infinite.

Answer 5

Ok, let's examine the equation behavior...

f(x) = x^2 sin(1/x)

x^2 goes to infinity as x goes to infinity

1/x goes to zero as x goes to infinity

sin(0) = 0.0

so, in the limit f(x) will converge to zero.

Answer 6

x^2 sin(1/x) goes to infinity as x tends to infinity

Answer 7

you want to adhere to l'hopitals rule you had the right mind-set: rewrite it as lim sin(a million/x) / (a million/x) now keep on with lhopital with the help of taking the spinoff of authentic and bottom (-cos(a million/x) / x²) / (-a million/x²) = cos(a million/x) now lim cos(a million/x) = cos (0) = a million

Answer 8

let x = 1/h
x->inf meand h->0
the value = 1/h^2(sin h)
= (sin h)/h^2

= ((sin h)/h)/h
as h->0 sinh/h->1
limit ->1/h or infinite