Question

pre calc help :)

Answer 1

Shoot

Answer 2

ok so i need to find the limit of the function algebraiaclly

Answer 3

lim {x^2-16/x-4}
x->5

Answer 4

factor the top

Answer 5

use the difference of squares

Answer 6

x-16/4

Answer 7

ohhhh

Answer 8

i see

Answer 9

x-4/1 ?

Answer 10

are you sure it is approaching 5, and not approaching 4?

Answer 11

you are correct 0.o

Answer 12

how did you know o.o

Answer 13

about what?

Answer 14

about the 4

Answer 15

because otherwise you can just plug it in.... it would be too simple for them to give it to you.

Answer 16

oh xD

Answer 17

Now however, \(\large\color{black}{ \displaystyle \lim_{x \rightarrow 4}~\frac{x^2-16}{x-4} }\) you need to factor and cancel things out, and only then plug 4 into the limit instead of x.

Answer 18

so if we factor we get x-4/1 correct

Answer 19

\(\large\color{black}{ \displaystyle \lim_{x \rightarrow 5}~\frac{x^2-16}{x-4}~\Longrightarrow~\lim_{x \rightarrow 5}~\frac{(x-4)(x+4)}{x-4} }\)

Answer 20

what cancels on top and bottom?

Answer 21

i canceled the x's

Answer 22

then simplified the 16/4

Answer 23

was that wrong o.o

Answer 24

\(\large\color{black}{ \displaystyle \lim_{x \rightarrow 4}~\frac{(x-4)(x+4)}{x-4} }\)

\(\large\color{black}{ \displaystyle \lim_{x \rightarrow 4}~\frac{\cancel{(x-4)}(x+4)}{\cancel{x-4}} }\)

\(\large\color{black}{ \displaystyle \lim_{x \rightarrow 4}~(x+4) }\)

Answer 25

then plug in 4 for x.

Answer 26

ohhhhhh k i see what you did

Answer 27

you split up the x^2 16 into 2 parenthesis xD

Answer 28

yes;)

Answer 29

and your final answer is?

Answer 30

and the final answer is 8 xD

Answer 31

yup, this limit is equivalent to 8. Correct!

Answer 32

thank you so much ^.^

Answer 33

yw