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

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Homework Help Overview

The problem involves finding the limit of the expression [arctan(x)-(x)] / [arcsin(x)-(x)] as x approaches 0, focusing on the behavior of these functions near zero.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to evaluate the limit by substituting values of x close to zero and observes the results, leading to confusion about the limit's existence and potential rounding issues with their calculator.

Discussion Status

Some participants suggest that the discrepancies in the calculated values are likely due to rounding errors, while others affirm the original suspicion that the limit is -2. The discussion reflects differing interpretations of the results and the reliability of the calculations.

Contextual Notes

Participants note the complexity of the arctan and arcsin functions and the limitations of handheld calculators in accurately computing values for small inputs.

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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.
 
Last edited:
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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
 
so did I compute it wrong? or is that just the way my calculator rounds it?
 
btw, thank you
 
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.
 
thank you for the help
 

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