Mathematics OpenStudy (anonymous):

need to attachment OpenStudy (anonymous): OpenStudy (anonymous):

what you wanna do here is multiply by the conjugate OpenStudy (anonymous):

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 OpenStudy (anonymous):

otherwise, no limit OpenStudy (anonymous):

so replace x by 0 and solve $\sqrt{b}-4=0$ $b=16$ OpenStudy (anonymous):

but how did you get the 16? OpenStudy (anonymous):

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 OpenStudy (anonymous):

oh because i want $\sqrt{b}-4=0$ $\sqrt{b}=4$ $b=16$ OpenStudy (anonymous):

oh ok you square both sides OpenStudy (anonymous):

well i guess. i actually just though "what number has square root 4?" OpenStudy (anonymous):

well that work to OpenStudy (anonymous):

you mean the conjugate which is the numerator and changing the plus to minus? OpenStudy (anonymous):

can someone please show me step by step how to multiply by the conjugate? OpenStudy (anonymous):

wow, you are late OpenStudy (anonymous):

yeah well my internet connection was messed up, lol OpenStudy (anonymous):

multiply top and bottom by $\sqrt{7x+16}+4$ numerator will be $7x+16-16=7x$ denominator will be $x(\sqrt{7x+16}+4)$ OpenStudy (anonymous):

fraction will reduce to $\frac{7}{\sqrt{x+16}+4}$ OpenStudy (anonymous):

now let x go to 0, get $\frac{7}{\sqrt{16}+4}=\frac{7}{8}$ OpenStudy (anonymous):

so is this when you would plug in the zero? OpenStudy (anonymous):

so is this the limit? OpenStudy (anonymous):

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

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