Question

lim(p→3)⁡(4x^2+5x-9)

Answer 1

there is content missing

Answer 2

lol, or its a convergence question ...

Answer 3

the limit of p as it approaches 3

Answer 4

this is correct

Answer 5

there is no p in your rule to assess it

Answer 6

lim(p→3)⁡(4x^2+5x-9) ; the value of p has no effect on this rule

Answer 7

hmm your right lol this must of been a mistake

Answer 8

lim(x→4)⁡(x^2-16)/(x-4)

Answer 9

how bout this one

Answer 10

thats more cogent :)

Answer 11

lets test something out, what is the value of the top and bottom when x = 4?

Answer 12

@Toadil do you know what a difference of squares is?

Answer 13

lim(x→4)⁡(4^2-16)/(0) undefined

Answer 14

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

Answer 15

since (x-4) is not factorable, then the top must have a factor of (x-4) hidden in it

Answer 16

okay

Answer 17

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

Answer 18

or some fancier new ones lol

Answer 19

(x^2-16) = 2(x - 8)

Answer 20

not quite, x^2 is not the same as 2x so that is not a proper application of factoring

Answer 21

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

Answer 22

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

Answer 23

so, what does this do for us?

\[\frac{x^2-16}{x-4}=\frac{\cancel{(x-4)}(x+4)}{\cancel{x-4}}=x+4\]

\[\lim_{x \to4}~(x+4)=(4+4)\]

Answer 24

As x approaches 4 x equals 4

Answer 25

as x approached 4, the function approaches 8

Answer 26

oh duh lol I was think x = 4

Answer 27

so the functions would equal 8 lol

Answer 28

4 + 4

Answer 29

graphically,

Answer 30

yes

Answer 31

the value of the function does not exist at x=4, but the limit exists as x approaches 4

Answer 32

Okay thank you very much for your help.

Answer 33

youre welcome