LIMITS approaches o+ - how come??

[noobie!]

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) hw cum i get negative infinity instead of positive infinity???it makes me so confuse..so could any1 plz rectify my mistakes for those theorem???please?! thanks alot 1st..:confused:

[arildno]

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

[noobie!]

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:

[adriank]

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

[noobie!]

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

thanks alot..rite nw i have no doubts..thanks for your kind help..thanks..:wink:

[noobie!]

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

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: