Limit of arctan(x)-x / arcsin(x)-x as x->0

  • Thread starter ggcheck
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In summary, the conversation discusses finding the limit of [arctan(x)-(x)] / [arcsin(x)-(x)] as x approaches 0. The individual attempted to calculate the limit by plugging in values of x closer and closer to 0, but encountered some rounding errors on their calculator. Ultimately, it is determined that the limit is -2, as originally suspected, and the discrepancies were due to the complexity of the functions and limitations of the calculator.
  • #1
ggcheck
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Homework Statement



I'm asked to find the limit of [arctan(x)-(x)] / [arcsin(x)-(x)] as x--->0

The Attempt at a Solution



So I started plugging in values of x closer and closer to zero. I get:

f(-0.05)= -1.994758
f(-0.01)= -1.999790
f(-0.001)= -2.000012

f(0.05)= -1.994758
f(0.01)= -1.999790
f(0.001)= -2.000024

At first (before I calculated (-0.001 and 0.001) I thought the limit was neg two, but after I calculated -.001 and .001 and got -2.000012 and -2.000024 I'm thrown off. Does the limit not exist, or is my calculator rounding wierdly.

Any help is much appreciated.
 
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  • #2
This is certainly a rounding error. The limit is -2 as u originally suspected. Here are some closer results:
f(0.001)=−1.995808383
f(-0.001)=−1.995808383
 
  • #3
so did I compute it wrong? or is that just the way my calculator rounds it?
 
  • #4
btw, thank you
 
  • #5
Arcsine and Arctan are relatively complex functions, and handheld calculators don't have the capability of doing a)operations with those complex numbers b)dividing them c)and worse of all, with really really small input values. So it is the way your calculator rounds.
 
  • #6
thank you for the help
 

What is the limit of arctan(x)-x / arcsin(x)-x as x->0?

The limit of arctan(x)-x / arcsin(x)-x as x->0 is equal to 1. This can be found by using L'Hopital's rule or by simplifying the expression using trigonometric identities.

How do I find the limit of arctan(x)-x / arcsin(x)-x as x->0?

To find the limit of arctan(x)-x / arcsin(x)-x as x->0, you can use L'Hopital's rule by taking the derivative of both the numerator and denominator. Alternatively, you can simplify the expression using trigonometric identities and then plug in x=0 to get the limit.

Why is the limit of arctan(x)-x / arcsin(x)-x as x->0 equal to 1?

The limit of arctan(x)-x / arcsin(x)-x as x->0 is equal to 1 because as x approaches 0, both the numerator and denominator approach 0. This results in an indeterminate form, which can be solved using L'Hopital's rule or by simplifying the expression to get the limit of 1.

Can the limit of arctan(x)-x / arcsin(x)-x as x->0 be evaluated without using L'Hopital's rule?

Yes, the limit of arctan(x)-x / arcsin(x)-x as x->0 can also be evaluated without using L'Hopital's rule. This can be done by simplifying the expression using trigonometric identities and then plugging in x=0 to get the limit of 1.

What is the significance of the limit of arctan(x)-x / arcsin(x)-x as x->0?

The limit of arctan(x)-x / arcsin(x)-x as x->0 is significant as it represents the behavior of the function as x approaches 0. It helps us understand the relationship between the two trigonometric functions and their behavior at the limit. This limit is also used in various mathematical and scientific applications, such as in the study of derivatives and integrals.

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