what is the value of Limit (x^2 - 16) / (x - 4) and x->4,if it exists?

2009-09-25 11:59 am
Find the value of Limit (x^2 - 16) / (x - 4) if it exists.

X -> 4

回答 (5)

2009-09-25 12:11 pm
✔ 最佳答案
x^2 - 16 = ( x - 4)( x + 4)

Hence, ( x^2 - 16 ) / ( x - 4 ) = x + 4.

Hence, the req'd limit is

= lim( x -> 4 ) ( x + 4 )

= 4 + 4

= 8. .....................................Ans.
2009-09-25 12:07 pm
(x^2 - 16) / (x - 4) = (x+4)(x-4) / (x-4) = x+4

= 8 as x->4
2016-12-04 12:46 pm
will we destroy x - sixteen down? we are in a position to if we cope with it via fact the version of two squares: x - sixteen => (sqrt(x) - 4) * (sqrt(x) + 4) (4 - sqrt(x)) / ((sqrt(x) - 4) * (sqrt(x) + 4)) => -a million / (sqrt(x) + 4) x is going to sixteen -a million / (sqrt(sixteen) + 4) => -a million / (+/- 4 + 4) => -a million / (-4 + 4) or -a million / (4 + 4) -a million / (-4 + 4) does not exist, so: -a million / 8 is the decrease
2009-09-25 12:08 pm
=Lim x-------->4 of (x-4)(x+4)/(x-4)
=Lim x-------->4 of (x+4)=8
God bless you.
2009-09-25 12:07 pm
lim(x→4) (x^2 - 16)/(x - 4)
= lim(x→4) (x^2 - 4^2)/(x - 4)
= lim(x→4) [(x + 4)(x - 4)]/(x - 4)
= lim(x→4) x + 4
= 4 + 4
= 8


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