Question

1) lim as x --> 0 (x^2 - 2x)/(sin3x)

Answer 1

I love your Question

Answer 2

omg are you from tennessee?!

Answer 3

what do you get to use?

Answer 4

\[\lim_{x \rightarrow 0 } \frac{x^2 -2x}{\sin 3x}\]
Apply L'Hospital If it is allowed

Answer 5

use l'hopital rule ===>

Answer 6

by which i really mean, can you use l'hopital or do you need something else?

Answer 7

We're supposed to just use the trig functions

Answer 8

okay If it is not allowed I still have way to do it

Answer 9

\[\lim_{x \rightarrow 0}\frac{x^2-2x}{\sin(3x)}=\lim_{x \rightarrow 0}\frac{x(x-2)}{\sin(3x)}=\frac{1}{3} \lim_{x \rightarrow 0}\frac{3x(x-2)}{\sin(3x)}\]
\[\frac{1}{3}\lim_{x \rightarrow 0} \frac{3x}{\sin(3x)} \cdot \lim_{x \rightarrow 0}(x-2)\]

Answer 10

\[=\frac{1}{3} \cdot 1 \cdot (0-2)\]

Answer 11

\[\lim_{x \rightarrow 0}\frac{x - 2 }{\frac{\sin3x}{x}*\frac{3}{3}}\]

Answer 12

\[\frac{1}{3}\]

Answer 13

\[=\frac{1}{3}(-2)=\frac{-2}{3}\]

Answer 14

\[\frac{-2}{3}\]Lol typo

Answer 15

Wow, thank you so much