Question

please help with limits

Answer 1

Find the limit of the function by using direct substitution.

limit as x approaches zero of quantity x squared plus four

Answer 2

so just plug in x=0 in that function

\(x^2+4 \\ =0^2+4 = ...\)

Answer 3

\[\lim_{x \rightarrow 0}(x^2+4) \]

Answer 4

so 4

Answer 5

yes :)

Answer 6

Find the limit of the function algebraically.

\[\lim_{x \rightarrow -4}(\frac{ x^2-16 }{ x+4 }) \]

Answer 7

so then i just put -4 in for these?

Answer 8

then you'll get a 0/0 which is an indeterminate form

Answer 9

so,
instead, factor out hthe numerator!

Answer 10

\(x^2-16=(...)(...)\)

Answer 11

okay so (x+4)(x-4)

Answer 12

correct!
what gets cancelled ?

Answer 13

\(\Large \dfrac{(x+4)(x-4)}{x+4}\)

Answer 14

\(\Large \dfrac{\cancel{(x+4)}(x-4)}{\cancel {x+4}} =...\)

Answer 15

after that you can directly plug in x = -4 :)

Answer 16

so its -8 c: thanks

Answer 17

yes, welcome ^_^