Question

\[lim_{x\to 0 }\frac{\tan x}{\sin 6x +\tan 7x}\]

Answer 1

divide numerator and denominator by tan x

Answer 2

break tan into sin/cos
for every cos you can directly put x=0
\(\dfrac{\sin m x}{\sin n x} = \dfrac{\dfrac{(m\cancel x)\sin mx}{mx}}{\dfrac{(n\cancel x)\sin nx}{nx}}\)

Answer 3

\[\huge \frac{1}{6x\frac{\sin 6x}{6x}\frac{\sin x}{x}\frac{x}{\cos x}+\frac{\tan7x}{\tan x}}\]

Answer 4

wait ithe cos is supposed to be on top and sin bottom hence the canceling

Answer 5

\[\lim_{x \to 0}\Large\frac{ 1 }{ \frac{ 6x\sin 6x }{ 6x } \frac{ \cos x }{ x}\frac{ x }{ \sin x }-\frac{\tan 7x}{\tan x}}\]

Answer 6

there are so many x's :O

Answer 7

sin 6x/ sin x will be just = 6, right ?

Answer 8

\(\dfrac{\sin m x}{\sin n x} = \dfrac{\dfrac{(m\cancel x)\sin mx}{mx}}{\dfrac{(n\cancel x)\sin nx}{nx}}= \dfrac{m}{n}\)

Answer 9

and sin 7x/ sin x is just 7
don't care about the cos's , they'll just be =1
1/(6+7)

Answer 10

its clear thank you ......@hartnn really nice trick

Answer 11

welcome ^_^