Limit as x goes to 5 from below

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Discussion Overview

The discussion centers around evaluating the limit of the expression $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$ as x approaches 5 from the left. Participants explore the implications of dividing by the highest power and the behavior of the numerator and denominator near this point.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes that dividing by the highest power leads to a situation where the numerator approaches $1-4/5$ while the denominator approaches $0$, questioning why this method does not yield a valid result.
  • Another participant points out that since the numerator is not zero and the denominator is zero at x = 5, this affects the limit calculation.
  • A participant asks how to calculate the limit given the previous observations.
  • One participant asserts that the limit does not exist as it approaches infinity.
  • Another participant acknowledges that while the overall limit does not exist, the left-hand limit can still be evaluated.
  • A follow-up question is raised regarding the implications of dividing by a very small number close to zero.
  • It is suggested that if the numerator is positive and the denominator is negative (close to zero), the limit approaches negative infinity.
  • A later reply confirms this interpretation, stating that the reasoning is correct.

Areas of Agreement / Disagreement

Participants generally agree that the limit does not exist, but there is a discussion about the left-hand limit and its evaluation. The interpretation of the behavior of the limit as it approaches negative infinity is also confirmed by some participants.

Contextual Notes

There are unresolved aspects regarding the method of dividing by the highest power and its implications for the limit, as well as the conditions under which the limit is evaluated.

Guest2
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Dividing by the highest power for $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$ I get

$\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}= \lim_{x \to 5^{-}}\frac{1-4/x}{1/x^{99}-5/x^{100}}$

However the denominator goes to $0$ whereas the numerator goes to $1-4/5$

Why isn't dividing by the highest power working?
 
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Guest said:
Dividing by the highest power for $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$ I get

$\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}= \lim_{x \to 5^{-}}\frac{1-4/x}{1/x^{99}-5/x^{100}}$

However the denominator goes to $0$ whereas the numerator goes to $1-4/5$

Why isn't dividing by the highest power working?

the reason in the given expression numerator is not zero but denominator is zero at x = 5.
hence after division also
 
kaliprasad said:
the reason in the given expression numerator is not zero but denominator is zero at x = 5.
hence after division also
So how does one calculate the limit?
 
Guest said:
So how does one calculate the limit?
\
so the limit does not exist as it goes to infinite
 
Kali, you are correct that the limit does not exist, but the LEFT HAND limit can still be evaluated.

To the OP - what happens when you divide by a very, very small number (close to 0)?
 
Prove It said:
Kali, you are correct that the limit does not exist, but the LEFT HAND limit can still be evaluated.

To the OP - what happens when you divide by a very, very small number (close to 0)?

you mean numerator is is positive and denominator is -ve( close to zero) so limit is - infinite ?
 
kaliprasad said:
you mean numerator is is positive and denominator is -ve( close to zero) so limit is - infinite ?

That is correct :)
 

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