LIMITS approaches o+ - how come?

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The discussion focuses on the evaluation of limits, specifically the behavior of the function \( \lim_{x \to -8^+} \frac{2x}{x+8} \). Participants confirm that as \( x \) approaches -8 from the right, the limit results in negative infinity due to the numerator being negative and the denominator being a small positive number. Additionally, the limit of the expression \( \left(\frac{1}{x^{1/3}} - \frac{1}{(x-1)^{4/3}}\right) \) as \( x \) approaches 0 from both sides is discussed, with the conclusion that it leads to positive infinity.

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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
 
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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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