LIMITS approaches o+ - how come?

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Discussion Overview

The discussion revolves around the concept of limits in calculus, specifically focusing on the behavior of functions as they approach certain values from the left and right. Participants explore the implications of approaching limits from positive and negative sides, particularly in the context of specific examples and homework problems.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant questions their understanding of limits, noting that as x approaches 0 from the positive side (0+), the limit tends to positive infinity, while approaching from the negative side (0-) tends to negative infinity.
  • Another participant prompts a consideration of the sign of a fraction when both the numerator and denominator are negative, suggesting that the result would be negative.
  • A participant explains that as x approaches -8 from the right, the function 2x/(x+8) approaches a large negative number due to the numerator being negative and the denominator being a small positive number.
  • Further, a participant expresses confusion about a different limit involving the expression [(1/x^(1/3)) - (1/(x-1)^(4/3))] as x approaches 0 from both sides, suggesting that the limit results in positive infinity but seeks clarification on this reasoning.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the behavior of limits as x approaches certain values, but there remains uncertainty and confusion regarding specific examples and the implications of signs in fractions. The discussion does not reach a consensus on all points raised.

Contextual Notes

Some participants express uncertainty about their understanding of limits and the behavior of functions near critical points, indicating potential gaps in foundational knowledge or specific mathematical steps that are not fully resolved.

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LIMITS approaches o+ - how come??

i encountered a few ques which makes me baffle whn i stdy about infinite limit.. as we know limit x --->0+ it will be positive infinity and when x-->o- it will be negative infinity; first of all m i rite?then..a que which i encountered was lim x --->-8+ (2x/ x+8) homework cum i get negative infinity instead of positive infinity?it makes me so confuse..so could any1 please rectify my mistakes for those theorem?please?! thanks a lot 1st..:confused:
 
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Well, what sign will a fraction between two negative numbers have? A negative sign, or a positive sign?
 


arildno said:
Well, what sign will a fraction between two negative numbers have? A negative sign, or a positive sign?

huh,umm i don't really get what u mean;but isit negative?:confused:
 


Intuitively, \lim_{x \to -8^+} 2x/(x+8) is what 2x/(x + 8) approaches as x approaches -8 from the right. If x is very slightly greater than -8, then 2x is negative (it's about -16), and x + 8 is a small positive number, so 2x/(x + 8) should be a large negative number. The graph below may help in visualizing what it looks like.

http://img380.imageshack.us/img380/7241/graphvn5.png
 
Last edited by a moderator:


adriank said:
Intuitively, \lim_{x \to -8^+} 2x/(x+8) is what 2x/(x + 8) approaches as x approaches -8 from the right. If x is very slightly greater than -8, then 2x is negative (it's about -16), and x + 8 is a small positive number, so 2x/(x + 8) should be a large negative number. The graph below may help in visualizing what it looks like.

http://img380.imageshack.us/img380/7241/graphvn5.png
[/URL]

thanks a lot..rite nw i have no doubts..thanks for your kind help..thanks..:wink:
 
Last edited by a moderator:


adriank said:
Intuitively, \lim_{x \to -8^+} 2x/(x+8) is what 2x/(x + 8) approaches as x approaches -8 from the right. If x is very slightly greater than -8, then 2x is negative (it's about -16), and x + 8 is a small positive number, so 2x/(x + 8) should be a large negative number. The graph below may help in visualizing what it looks like.

http://img380.imageshack.us/img380/7241/graphvn5.png
[/URL]

one more doubt is let say an example: [(1/x^1/3) - (1/(x-1)^4/3 ] ;its limit is x --->0+ and 0- so the answer will be positive infinity because of v sub x=o into the equation ,thus its infinity minus 3 that's why we got positive infinity same goes to 0- ?please rectify my mistakes if i have thm..thanks :blushing:
 
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