# Limit of Arcsin Approaching Infinity

• tsw303
In summary, the limit of (arcsin(x))/x as x approaches infinity is equal to 0. This is determined by using the sandwich theorem and observing the graph of (arcsin(x))/x. However, this may only be true for complex limits as arcsin(x) is only defined for x between -1 and 1 in the real numbers. Using L'Hopital's Rule may provide a complex answer, but it is not applicable in this case. Therefore, it is possible that this limit does not exist for real numbers.

## Homework Statement

limit approaching infinity: (arcsin(x))/(x)

= 0

Question is: Why? The 'Sandwich Theorem' 0=[(arcsinx)/x]=0 gives this
solution, but looking at the graph of (arcsinx)/x , this appears
impossible.

## Homework Equations

lim x->OO [arcsin(x)] - {DNE)
lim x->OO [(arcsinx)/x] - {DNE/OO}

lim x->OO [-Pi/2x] = lim x->OO [(arcsinx)/x] = lim x->OO [Pi/2x]
=> 0 = [(arcsinx)/x] = 0

=> lim x->OO [(arcsinx)/x] = 0

HOWEVER: (-Pi/2)<[arcsinx]<(Pi/2) ... SO ...
... (x) never reaches (OO) for (Pi/2x) to reach the limit x->OO (Pi/OO)
= 0.

## The Attempt at a Solution

The sandwich theorem gives and answer of "0". Maple 11 gives the same answer, so does my teacher. This just doesn't seem possible.

The only way you can get "0" is if you're using complex limits. I'm pretty sure. Let me check up on that.

Last edited:
If you use, L'Hospitals Rule, you get a $$\frac{1}{\sqrt{1-x^2}}$$, for x tending to infinity. Thats a zero, but a complex one, don't know if its even applicable here...

arcsin(x) isn't defined for x outside of [-1, 1] considering only the reals. So unless x is complex, I don't think this limit exists.

## 1. What is the limit of Arcsin(x) as x approaches infinity?

The limit of Arcsin(x) as x approaches infinity is undefined, or does not exist. This is because the inverse sine function, Arcsin(x), can only take values between -π/2 and π/2, and as x approaches infinity, the values of Arcsin(x) become increasingly large and undefined.

## 2. Can the limit of Arcsin(x) as x approaches infinity be evaluated using L'Hopital's rule?

No, L'Hopital's rule cannot be used to evaluate the limit of Arcsin(x) as x approaches infinity. This is because L'Hopital's rule can only be applied to indeterminate forms, and the limit of Arcsin(x) as x approaches infinity is not an indeterminate form.

## 3. Is there a way to express the limit of Arcsin(x) as x approaches infinity using other mathematical functions?

Yes, the limit of Arcsin(x) as x approaches infinity can be expressed using the inverse tangent function, Arctan(x). The limit can be rewritten as Arctan(1/x) as x approaches 0 from the positive side.

## 4. What is the graphical representation of the limit of Arcsin(x) as x approaches infinity?

The graphical representation of the limit of Arcsin(x) as x approaches infinity is a horizontal asymptote at y = π/2. This means that as x approaches infinity, the graph of Arcsin(x) approaches the line y = π/2 but never touches it.

## 5. How does the limit of Arcsin(x) as x approaches infinity relate to the concept of inverse trigonometric functions?

The limit of Arcsin(x) as x approaches infinity is related to the concept of inverse trigonometric functions by showing that the inverse sine function, Arcsin(x), has a limited range of values and is undefined for certain input values. This highlights the importance of understanding the domain and range of inverse trigonometric functions in order to properly evaluate their limits.