Calculating Limits of Sin(1/x)/(1/x)

[seto6]

Homework Statement

ok, this is the limit i want to calculate: limit x->0 sin(1/x) / (1/x) and

limit x->infinity sin(1/x) / (1/x)

can you tell me the clear step mathematical proof
2. The attempt at a solution

now the first one i tired doing this, let u =(1/x) then: limit U->infinity sin(U) / (U)

now i understand its actually zero since, sin(U) goes back between -1 and 1, when u grows to infinity so its zero...so what's the mathematical proof.

 

[Char. Limit]

You can't take the limit as x goes to infinity of sin(u)/u...

You need to find something else.

As x goes to infinity, what does u go to?

Because that's the limit you need.

 

[seto6]

You can't take the limit as x goes to infinity of sin(u)/u...

You need to find something else.

As x goes to infinity, what does u go to?

Because that's the limit you need.

My mistake it was suppose to be u, since u=1/x...as x->0...u->infinity

 

[Char. Limit]

My mistake it was suppose to be u, since u=1/x...as x->0...u->infinity

But in the problem you were working on last, x wasn't going to zero. x was going to infinity.

However, yes, that will be useful in the second problem.

 

[Hoblitz]

I can tell for the first one you're nearly there... after you've rewritten lim (x to 0) sin(1/x) / (1 / x) with u = 1/x, we have lim (u to infinity) sin (u) / u.

In this case, how does the Squeeze theorem apply?

For the second case, you might consider L' Hospital's rule, or the power series expansion of sine.

 

[seto6]

for the second case:

limit x->infinity sin(1/x) / (1/x) if i appy l'hospital...

limit x->infinity (-1/x^2) (cos(1/x)) / (-1/x^2)

thus i get limit x->infinity cos(1/x) and again stuck.

 

[irycio]

Why are you stuck? If x goes to infinity, then 1/x goes to 0, and cos(0) is well defined :)