Calculus, L'Hopital's Rule?

Question

Hi, I was wondering how I should do this problem and if I am doing it correctly.

lim x -> +infinity
xSIN(1/x)
= xSINx^-1
= -COSx^-2
= 1/COSx^2
Im not sure what to do after this or if I did this right.
Thanks for the help

Answer 1

Lim(x->oo)[xSin[1/x]]=
Lim(x->oo)[Sin[1/x]/(1/x)]=0/0
so now we can use l'hopital's rule
the limit becomes:
Lim(x->oo)
[(-1/x^2)Cos[1/x]/(-1/x^2)]=
Lim(x->oo)[Cos[1/x]]=Cos[0]=1

Answer 2

I have never seen L'Hopital's rule applied to an infinite limit, but there is an obvious transformation that would make it applicable. Let z = 1/x. Then, we want the limit, as z approaches 0, of: (1/z) sin z, or (sin z)/z. This is clearly bait for L'Hopital, and we wind up with cos z/1 at z = 0. The answer is obviously 1.

Answer 3

you problem is , you apply L'Hopital Rules wrongly . L'Hopital rules must have the pattern

infinity/infinity

OR

0/0

So you have to convert them into such pattern first .

But there is another easier way to do it . Try the substitution 1/x = y . So when x approaches infinity , y approaches 0 and your question become

Lim y->0 (sin y)/y = 1

Answer 4

I'm not quite sure what you've done there; a few brackets might help :)
Remember you need a fraction:
x sin(1/x) = sin(1/x) / (1/x).
Differentiate top and bottom:
[cos (1/x) * -1/x^2]/[-1/x^2]
= cos(1/x).
Now, as x goes to infinity, cos(1/x) goes to cos(0) = 1.
So the answer is 1.

Answer 5

rearrange to lim x ->infinity sinx/(1/x),

let u = 1/x, the equation become lim u->0 sinu/u, apply L'Hopital rule the limit is 1

Answer 6

I would rewrite as
f(x)/g(x) where f(x) = sin(1/x) and g(x) = (1/x)
Both numerator and denominator approach zero as x approaches infinity.

The derivatives are
f'(x) = cos(1/x)*(-1/x^2)
and
g'(x) = -1/x^2
both derivatives approach zero.

So L'Hospital's rule should work.
f'(x)/g'(x) = cos(1/x)*(-1/x^2)/(-1/x^2) = cos(1/x)

And limit(cos(1/x)) as x --> infinity is cos(0) = 1
I think you made a mistake in taking the derivatives