Lim x->0 (sin x / x) =1 contradiction?

[khurram usman]

lim x->0 (sin x / x) =1...contradiction?

sin(x)/x =1 (limit x to 0)
this is an identity proved by using geometry and squeeze theorem ...right?

now today i came across another question and doing it my way ...gives me two answers;)
the question is limit x-->0 of [ x*sin(1/x)]

my first approach was using the above identity by rewriting the question as follows:
sin(1/x)/(1/x)...it means the same thing and is now in the form so that we can use the identity...so limiting x->0 must give us 1 according to the identity

now i thought to use the squeeze theorem as shown below:
-1≤sin(1/x)≤1
-x≤x*sin(1/x)≤x
now x goes to 0 so:
0≤x*sin(1/x)≤0
so x*sin(1/x)=0 as x goes to 0

now which method which is correct ?

 

[disregardthat]

sin(1/x)/(1/x) as x goes to 0 would be the same as sin(x)/x as x goes to infinity, not 0.

 

[khurram usman]

sin(1/x)/(1/x) as x goes to 0 would be the same as sin(x)/x as x goes to infinity, not 0.

so it will be of the form sin(x)/x
applying l'hospital's rule :
cos(x)/1
= cos(x)
as you said x goes to infinity so what will be cos(∞)?

 

[D H]

so it will be of the form sin(x)/x
applying l'hospital's rule :
cos(x)/1
= cos(x)
as you said x goes to infinity so what will be cos(∞)?

It's undefined, and all that that means is that you can't use l'Hôpital's rule here.

You can use the squeeze theorem on sin(x)/x as x→∞. sin(x) is bounded from above by +1, from below by -1. Given that, what does the squeeze theorem say about sin(x)/x as x→∞?

 

[khurram usman]

It's undefined, and all that that means is that you can't use l'Hôpital's rule here.

You can use the squeeze theorem on sin(x)/x as x→∞. sin(x) is bounded from above by +1, from below by -1. Given that, what does the squeeze theorem say about sin(x)/x as x→∞?

ok ...got your point...its 0
thx a lot

 

[mathman]

so it will be of the form sin(x)/x
applying l'hospital's rule :
cos(x)/1
= cos(x)
as you said x goes to infinity so what will be cos(∞)?

L'Hopital's rule doesn't apply, since |sin(x)| is bounded while the denominator x becomes infinite.