Why is 1^∞ an Indeterminate Form?

  • Thread starter mbrmbrg
  • Start date
  • Tags
    Forms
In summary: But this is not a rigorous definition, and I'm not sure if my explanation is correct enough to be useful.In summary, 1^\infty is an indeterminate form because when taking the limit of a function involving 1 raised to a variable that goes to infinity, the result depends on the speed of convergence of the function. This can be seen in examples where the limit may converge to 1, 0, or another value depending on the behavior of the function at infinity. The concept of speed convergence is often used to explain this phenomenon, but a rigorous definition is still debatable.
  • #1
mbrmbrg
496
2
Why is [tex]1^\infty[/tex] an indeterminate form? If you keep multiplying 1 by itself doesn't the answer stay 1?
 
Physics news on Phys.org
  • #2
It is an indeterminate form because [tex]0 . \infty[/tex] is.

Try to take the "logarithm of [tex]1^\infty[/tex]".

EDIT:
Obviously it is not a proof.
It is only a mnemonic trick.
 
Last edited:
  • #3
Sure,

[tex]\lim_{x\rightarrow \infty} 1^x=1[/tex].

But consider a case like

[tex]\lim_{x\rightarrow \infty} (1+\frac{1}{x})^{e^{x^5}}[/tex]

1+1/x goes to 1, but does it so converges much slower than exp(x^5) goes to infinity! So if you think about it this way, maybe you see that it makes sense that sometimes a limit of the "form" [itex]1^{\infty}[/tex] will diverge, and some other times, it will behave and go to 1.

Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?
 
Last edited:
  • #4
quasar987 said:
Sure,

[tex]\lim_{x\rightarrow \infty} 1^x=1[/tex].

And, obviously,

[tex]\lim_{x\rightarrow \infty} 0 x = 0[/tex]


Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?

[tex]\lim_{x\rightarrow \infty} (\cos(1/x^2))^{x^4}[/tex]
 
Last edited:
  • #5
quasar987 said:
Sure,

[tex]\lim_{x\rightarrow \infty} 1^x=1[/tex].

But consider a case like

[tex]\lim_{x\rightarrow \infty} (1+\frac{1}{x})^{e^{x^5}}[/tex]

1+1/x goes to 1, but does it so converges much slower than exp(x^5) goes to infinity! So if you think about it this way, maybe you see that it makes sense that sometimes a limit of the "form" [itex]1^{\infty}[/tex] will diverge, and some other times, it will behave and go to 1.

Maybe in some cases, the limit will converge to a value that is not 1? Can someone provide an exemple?

Yeah, that's what I was talking about. I know that the form [tex]1^{\infty}[/tex] has many different values; I wanted to know why.
Can you run through the "converging at different speeds" bit a lot slower (pun unintended, but noted and pride taken)?
 
  • #6
I started writing an explanation of the concept of speed convergence, but in the middle of it, I realized that I'm too unsure about too many points, and the explanation is unnecessarily too blurry imo. So I'll let someone more qualified answer.

Informally, the concept of speed convergence is incarnated in L'Hospital's rule: if both f and g go to infinity, then the limit of their ratio is determined by the ratio of their speed (derivative) at infinity.
 
Last edited:

FAQ: Why is 1^∞ an Indeterminate Form?

What is an indeterminate form?

An indeterminate form is a mathematical expression that cannot be evaluated to a specific value using basic arithmetic operations. These forms typically involve the division of two quantities that approach infinity or zero.

Why is 1^∞ considered an indeterminate form?

When we take the limit of 1 raised to the power of infinity, the result is undefined because we do not know whether the exponent or the base is increasing faster. This uncertainty leads to an indeterminate form.

Can we manipulate 1^∞ to be evaluated to a specific value?

No, 1^∞ will always be an indeterminate form. However, we can use other mathematical techniques, such as L'Hospital's rule or logarithms, to evaluate the limit and determine its value.

What are some real-world applications of 1^∞ as an indeterminate form?

Indeterminate forms, including 1^∞, are commonly used in calculus and other branches of mathematics. They can also be applied in physics, particularly in the study of limits and infinitesimals.

Is 1^∞ always considered an indeterminate form?

No, there are certain cases where the limit of 1^∞ can be evaluated to a specific value. For example, if the exponent is an odd integer, the limit will approach either 1 or negative 1 depending on the sign of the base. However, in general, 1^∞ is considered an indeterminate form.

Back
Top