need to attachment

what you wanna do here is multiply by the conjugate

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

otherwise, no limit

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

but how did you get the 16?

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

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

oh ok you square both sides

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

well that work to

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

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

wow, you are late

yeah well my internet connection was messed up, lol

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

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

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

so is this when you would plug in the zero?

so is this the limit?

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

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