anonymous
  • anonymous
lim(p→3)⁡(4x^2+5x-9)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
there is content missing
amistre64
  • amistre64
lol, or its a convergence question ...
anonymous
  • anonymous
the limit of p as it approaches 3

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anonymous
  • anonymous
this is correct
amistre64
  • amistre64
there is no p in your rule to assess it
amistre64
  • amistre64
lim(p→3)⁡(4x^2+5x-9) ; the value of p has no effect on this rule
anonymous
  • anonymous
hmm your right lol this must of been a mistake
anonymous
  • anonymous
lim(x→4)⁡(x^2-16)/(x-4)
anonymous
  • anonymous
how bout this one
amistre64
  • amistre64
thats more cogent :)
amistre64
  • amistre64
lets test something out, what is the value of the top and bottom when x = 4?
anonymous
  • anonymous
@Toadil do you know what a difference of squares is?
anonymous
  • anonymous
lim(x→4)⁡(4^2-16)/(0) undefined
amistre64
  • amistre64
4^2 - 16 = 0 4 - 4 = 0 since both equate to zero, this tell us that they have a common factor get them into their factored form to cancel out the like factor and reassess it
amistre64
  • amistre64
since (x-4) is not factorable, then the top must have a factor of (x-4) hidden in it
anonymous
  • anonymous
okay
amistre64
  • amistre64
are you able to factor the top? oldrin suggested its a difference of squares if that rings a bell; if not, then you would have to rely upon a more basic method
amistre64
  • amistre64
or some fancier new ones lol
anonymous
  • anonymous
(x^2-16) = 2(x - 8)
amistre64
  • amistre64
not quite, x^2 is not the same as 2x so that is not a proper application of factoring
amistre64
  • amistre64
we know that (x-4) is a factor (x-4)(x+n) = x^2-16 , distribute the left side (x-4)x+(x-4)n = x^2-16 , distribute the left side again xx-4x+nx-4n = x^2-16 , add 0x to the right side for comparison x^2 +(n-4)x -4n = x^2 + 0x -16 , and compare parts x^2 +(n-4)x -4n x^2 + 0 x -16 n-4 = 0 and -4n = 16 solve for n
amistre64
  • amistre64
if we go back to grade school and just divide it out ... x + 4 <-- our other factor ------------- x-4 | x^2 - 16 -(x^2 -4x) ---------- 4x - 16 -(4x - 16) --------- 0 + 0 <-- nothing remains therefore; (x-4)(x+4) = x^2 - 16
amistre64
  • amistre64
so, what does this do for us? \[\frac{x^2-16}{x-4}=\frac{\cancel{(x-4)}(x+4)}{\cancel{x-4}}=x+4\] \[\lim_{x \to4}~(x+4)=(4+4)\]
anonymous
  • anonymous
As x approaches 4 x equals 4
amistre64
  • amistre64
as x approached 4, the function approaches 8
anonymous
  • anonymous
oh duh lol I was think x = 4
anonymous
  • anonymous
so the functions would equal 8 lol
anonymous
  • anonymous
4 + 4
amistre64
  • amistre64
graphically,|dw:1378656495602:dw|
amistre64
  • amistre64
yes
amistre64
  • amistre64
the value of the function does not exist at x=4, but the limit exists as x approaches 4
anonymous
  • anonymous
Okay thank you very much for your help.
amistre64
  • amistre64
youre welcome

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