Question

need to attachment

Answer 1

what you wanna do here is multiply by the conjugate

Answer 2

ok idea is this. if the limit is going to exist, since the denominator is going to 0 the numerator must go to 0 as well

Answer 3

otherwise, no limit

Answer 4

so replace x by 0 and solve
\[\sqrt{b}-4=0\]
\[b=16\]

Answer 5

but how did you get the 16?

Answer 6

now as lagrange said, if you want to find the limit of
\[\lim_{x\rightarrow 0}\frac{\sqrt{7x+16}-4}{x}\] you have to multiply by the conjugate

Answer 7

oh because i want
\[\sqrt{b}-4=0\]
\[\sqrt{b}=4\]
\[b=16\]

Answer 8

oh ok you square both sides

Answer 9

well i guess. i actually just though "what number has square root 4?"

Answer 10

well that work to

Answer 11

you mean the conjugate which is the numerator and changing the plus to minus?

Answer 12

can someone please show me step by step how to multiply by the conjugate?

Answer 13

wow, you are late

Answer 14

yeah well my internet connection was messed up, lol

Answer 15

multiply top and bottom by
\[\sqrt{7x+16}+4\]
numerator will be
\[7x+16-16=7x\]
denominator will be
\[x(\sqrt{7x+16}+4)\]

Answer 16

fraction will reduce to
\[\frac{7}{\sqrt{x+16}+4}\]

Answer 17

now let x go to 0, get
\[\frac{7}{\sqrt{16}+4}=\frac{7}{8}\]

Answer 18

so is this when you would plug in the zero?

Answer 19

so is this the limit?

Answer 20

yes you plug in 0 when it is no longer in the denominator