Question

Q: the limit as x approaches 2 for the square root of 4x+1,-3, all over x-2

Answer 1

\[\lim_{x \rightarrow 2}\frac{-3 \sqrt{4x+1}}{x-2}\]

Answer 2

but honestly i don't know what i'm reading

Answer 3

\[\lim_{x \rightarrow 2}(\sqrt{4x+1})\div(x-2)\]

Answer 4

\[\lim_{x ->2}\frac{ \sqrt{4x+1}-3}{x-2}\]

Answer 5

so just ignore -3?

Answer 6

usually a sqrt tends to have you play congutes

Answer 7

conjugates

Answer 8

\[\lim_{x \rightarrow 2}\frac{\sqrt{4x+1}-3}{x-2} \cdot \frac{\sqrt{4x+1}+3}{\sqrt{4x+1}+3}\]

Answer 9

\[\lim_{x \rightarrow 2}\frac{(4x+1)-9}{(x-2)(\sqrt{4x+1}+3)}\]

Answer 10

if it's amistre's one.....then I will use l'Hospital

Answer 11

\[\lim_{x \rightarrow 2}\frac{4x-8}{(x-2)(\sqrt{4x+1}+3)}=\lim_{x \rightarrow 2}\frac{4(x-2)}{(x-2)(\sqrt{4x+1}+3)}\]

Answer 12

\[\lim_{x \rightarrow 2}\frac{4}{\sqrt{4x+1}+3}=\frac{4}{\sqrt{4(2)+1}+3}\]

Answer 13

\[\lim_{x \rightarrow 2} \frac{4}{\sqrt{4x+1}\times1}\]
\[\frac{4}{3}\]

Answer 14

\[=\frac{4}{\sqrt{8+1}+3}=\frac{4}{\sqrt{9}+3}=\frac{4}{3+3}=\frac{4}{6}\]

Answer 15

umm I messed up somewhere maybe

Answer 16

ah Yeah I did in differentiation myininaya is right

Answer 17

\[=\frac{2 \cdot 2}{2 \cdot 3}=\frac{2}{3}\]

Answer 18

\[\lim_{x ->2}\frac{Dx( \sqrt{4x+1}-3)}{Dx(x-2)}\]
\[\lim_{x ->2}\frac{Dx( \sqrt{4x+1}-3)}{1}\]
\[\lim_{x ->2}Dx( \sqrt{4x+1}-3)\]
\[\lim_{x ->2}\left( \frac{Dx(4x-1)}{2\sqrt{4x+1}}\right)\]
\[\lim_{x ->2}\left( \frac{4}{2\sqrt{4x+1}}\right)\]
\[ \frac{4}{2\sqrt{4(2)+1}}\]
\[ \frac{2}{\sqrt{9}}= \frac{2}{3}\]

Answer 19

per ishaan :)