Question

Evaluate the limit:
lim (-7x)/(sqrt(4x+16)-4) where x approaches 0

Answer 1

i think it goes to -14 i used l'hospitals rule cuz the fist part is 0/0

Answer 2

Can you show me how you did that?

Answer 3

1.without L'hospitals rule \[\lim_{x\rightarrow 0}\frac{-7x}{\sqrt{4x+16}-4}=\lim_{x\rightarrow 0}\frac{-7x}{\sqrt{4x+16}-4}\cdot\frac{\sqrt{4x+16}+4}{\sqrt{4x+16}+4}=\]\[=\lim_{x\rightarrow 0}\frac{-7x(\sqrt{4x+16}+4)}{4x+16-16}=\lim_{x\rightarrow 0}\frac{-7x(\sqrt{4x+16}+4)}{4x}=\]\[=\lim_{x\rightarrow 0}\frac{-7(\sqrt{4x+16}+4)}{4}=-14\]

Answer 4

2.with L'hospitals rule \[\lim_{x\rightarrow 0}\frac{-7x}{\sqrt{4x+16}-4}=\lim_{x\rightarrow 0}\frac{(-7x)'}{(\sqrt{4x+16}-4)'}=\]\[=\lim_{x\rightarrow 0}\frac{-7}{2/\sqrt{4x+16}}=\lim_{x\rightarrow 0}\frac{-7\sqrt{4x+16}}{2}=-14\]