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

  • Thread starter seto6
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In summary: The Squeeze theorem applies because you're trying to take the limit as x goes to infinity of sin(u)/u...If you use L'Hospital's rule, you'll get the same answer. If you use the power series expansion of sin, you'll get cos(1/x) and be stuck again.
  • #1
seto6
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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.
 
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  • #2
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.
 
  • #3
Char. Limit said:
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
 
  • #4
seto6 said:
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
Why are you stuck? If x goes to infinity, then 1/x goes to 0, and cos(0) is well defined :)
 

1. What is the formula for calculating limits of sin(1/x)/(1/x)?

The formula for calculating the limit of sin(1/x)/(1/x) is: limx→0 sin(1/x)/(1/x).

2. How do I find the limit of sin(1/x)/(1/x)?

To find the limit of sin(1/x)/(1/x), you can plug in increasingly smaller values for x, such as 0.1, 0.01, and 0.001, and observe the result. If the result approaches a specific value (known as the limit), then that is the limit of the function.

3. Can I use L'Hôpital's rule to find the limit of sin(1/x)/(1/x)?

Yes, L'Hôpital's rule can be used to find the limit of sin(1/x)/(1/x). The rule states that if the limit of a function is of the form 0/0 or ∞/∞, then it can be found by taking the derivative of both the numerator and denominator and then evaluating the limit again.

4. What is the value of the limit of sin(1/x)/(1/x)?

The value of the limit of sin(1/x)/(1/x) is undefined. As x approaches 0, the function oscillates between -1 and 1, never settling on a specific value. This is known as an oscillating or indeterminate limit.

5. Why is calculating the limit of sin(1/x)/(1/x) important in mathematics?

Calculating limits is important in mathematics because it allows us to determine the behavior of a function as the input approaches a certain value. In the case of sin(1/x)/(1/x), the limit helps us understand how the function behaves near 0, which can have implications in areas such as trigonometry and calculus.

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