Solve Quantum Physics Limit: tan(ax)/x → 0 as x→∞

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In summary, maria clara suggested that boombaby was right and that tanh does approach zero as x gets larger. Furthermore, as maria clara found out, it was all her mistake and tanh does in fact approach zero as x gets larger.
  • #1
maria clara
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Hi, I've been trying to solve a problem in quantum physics, and got stuck because of a limit. I guess I'm a little rusty on that and would appreciate any help.
How can I show that the expression tan(ax)/x tends to zero in the limit x---> infinity?
thanks!
 
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  • #2
tan(ax)=sin(ax)/cos(ax)

thus tan(ax)/x=sin(ax)/xcos(ax)

as x-> infinity, does sin(ax) and cos(ax) approach any single value?
If it does, then you can find the answer easily

If it doesn't, then your limit would just depend on the 'x' in the denominator if you understand what I am saying.
 
  • #3
it does not converge
 
  • #4
well, that's exactly the problem, sin and cos don't approach a certain value at infinity, and 1/x does. But is there a theorem that states that if a function approches zero and another function does not approach any specific value, then the product of both would approach zero? I don't think so, there has to be a better explanation. If I'm wrong, please correct me, thanks again!
 
  • #5
maria clara said:
Hi, I've been trying to solve a problem in quantum physics, and got stuck because of a limit. I guess I'm a little rusty on that and would appreciate any help.
How can I show that the expression tan(ax)/x tends to zero in the limit x---> infinity?
thanks!

Hi maria clara! :smile:

I think boombaby is right …

tan keeps hitting ±∞ every π/a …

can you find an N such that, for x > N, |tan(ax)/x| is always < 1, for examle?
 
  • #6
you're right, it doesn't converge.

and I just found out that it was all my mistake, it was tanh and not tan... and since tanh is bounded at infinity, tanh(ax)/x definitely approaches zero...

sorry guys.. thanks for your help anyway...:)
 

1. What is the significance of the limit tan(ax)/x as x approaches infinity in quantum physics?

The limit tan(ax)/x as x approaches infinity is an important concept in quantum physics as it helps to determine the behavior of certain particles at extremely high energies. It is used to study the properties of particles such as photons and electrons and their interactions with electromagnetic fields.

2. How is the limit tan(ax)/x evaluated in quantum physics?

The limit tan(ax)/x is evaluated by using mathematical techniques such as L'Hopital's rule, which allows for the calculation of limits involving trigonometric functions. It is also evaluated using the principles of quantum mechanics, which describe the behavior of particles at the quantum level.

3. What does it mean if the limit tan(ax)/x approaches 0 as x becomes infinitely large in quantum physics?

If the limit tan(ax)/x approaches 0 as x becomes infinitely large, it means that the function approaches a value of 0 at infinity. This indicates that the behavior of the particles being studied is becoming more regular and predictable at higher energies.

4. Can the limit tan(ax)/x ever equal 0 in quantum physics?

No, the limit tan(ax)/x can never equal 0 in quantum physics as it is an asymptotic limit. This means that the function can approach 0, but can never actually reach it. It is a representation of the behavior of particles at extremely high energies, but does not have a physical value itself.

5. How does the limit tan(ax)/x relate to Heisenberg's uncertainty principle in quantum physics?

The limit tan(ax)/x relates to Heisenberg's uncertainty principle in that it represents the behavior of particles at energies where the uncertainty principle becomes significant. The uncertainty principle states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. The limit tan(ax)/x helps to understand this behavior at extremely high energies.

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