Question

lim as x --> 0 ((sin3x)(sin2x))/(xsin5x)

Answer 1

still no l'hopital, lots of algebra. answer is
\[\frac{6}{5}\] myininaya will write it out

Answer 2

i am anxiously awaiting this algebra. also a reasonable explanation of how you decide what to do

Answer 3

likewise! thanks for the help.

Answer 4

(only saying that because i have no idea how to do it)

Answer 5

first of all we know \[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \] by squeeze thm
now let's prove something else
\[\lim_{x \rightarrow 0}\frac{\sin(Ax)}{\sin(Bx)}=\frac{A}{B}\]
\[\lim_{x \rightarrow 0}\frac{\sin(Ax)}{\sin(Bx)} \cdot \frac{ABx}{ABx}=\lim_{x \rightarrow 0}\frac{A}{B} \cdot \frac{\sin(Ax)}{Ax} \cdot \frac{Bx}{\sin(Bx)}=\frac{A}{B} \cdot 1 \cdot 1=\frac{A}{B}\]
so we have
\[\lim_{x \rightarrow 0}\frac{\sin(3x)}{x} \cdot \frac{\sin(2x)}{\sin(5x)} \cdot \frac{3}{3}\]
\[=3\lim_{x \rightarrow 0}\frac{\sin(3x)}{3x} \cdot \frac{\sin(2x)}{\sin(5x)}=3 \cdot 1 \cdot \frac{2}{5}\]

Answer 6

very nice explanation!

Answer 7

:)

Answer 8

awesome. you blew my mind!

Answer 9

myininaya is the limit!