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LIMITS approaches o+ - how come?

  1. Nov 29, 2008 #1
    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) 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:
     
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  3. Nov 29, 2008 #2

    arildno

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    Re: LIMITS approaches o+ - how come??

    Well, what sign will a fraction between two negative numbers have? A negative sign, or a positive sign?
     
  4. Nov 29, 2008 #3
    Re: LIMITS approaches o+ - how come??

    huh,umm i don't really get what u mean;but isit negative?:confused:
     
  5. Nov 29, 2008 #4
    Re: LIMITS approaches o+ - how come??

    Intuitively, [tex]\lim_{x \to -8^+} 2x/(x+8)[/tex] 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.

    [​IMG]
     
  6. Nov 30, 2008 #5
    Re: LIMITS approaches o+ - how come??

    thanks alot..rite nw i have no doubts..thanks for your kind help..thanks..:wink:
     
  7. Nov 30, 2008 #6
    Re: LIMITS approaches o+ - how come??

    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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