Improper Integral: Limit Can Be Zero Explained

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Homework Help Overview

The discussion revolves around understanding the behavior of an improper integral and the application of limits, particularly in the context of L'Hospital's rule and the evaluation of limits that may approach zero.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are examining the application of L'Hospital's rule and questioning the correctness of the product rule in the context of limits. There is also a discussion about the behavior of specific functions as they approach zero.

Discussion Status

The conversation includes various interpretations of the limit in question, with some participants asserting that the limit approaches zero while others express confusion about the application of rules. Guidance has been offered regarding the bounded nature of certain functions involved.

Contextual Notes

There appears to be a lack of clarity regarding the steps taken in the limit evaluation, as well as differing opinions on the correctness of the approaches used. Some participants note the need for more explicit detail in the calculations presented.

SANGHERA.JAS
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How the following limit can be zero, since after applying L'Hospital rule the root x will be in the numerator, which together with sinx will constitute infinity.
 

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I'm not sure you are actually using the product rule correctly. Hard to tell since you don't show what you actually did. But you have an x in the denominator. It's still infinity/infinity.
 
The proceedure is following, In my book they have equate it to zero.
 

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Your limit is completely wrong, the limit is 0. (1-cos(x))/x^1/2 goes to zero. Why do you think it doesn't? (1-cos(x)) is bounded. x^(1/2) isn't.
 
Oh, I got it. Thanks a lot.
 

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