trig limits help

[solid sequoia]

I got to here

[nocturne arch]

,,9{\color{yellow}\tan x}=9{\color{yellow}\frac{\sin x}{\cos x}}=\frac{9\sin x}{\cos x}\text{, not }\frac{9\sin x}{9\cos x}

[pearl spade]

mtt07734

[nocturne arch]

also $\frac{\sin x}{}\cdot\frac{}{\sin x}$ cancel out

[pearl spade]

mtt07734

[nocturne arch]

so instead you should have:

,,\frac{9\tan x}{3\sin x}=\frac{\frac{9\sin x}{\cos x}}{3\sin x}=\frac{9\sin x}{\cos x}\cdot\frac1{3\sin x}=\frac9{\cos x}\cdot\frac13

[pearl spade]

mtt07734

[nocturne arch]

then with cos(0) = 1, you can get the correct answer for the limit

[solid sequoia]

Ah ok

[solid sequoia]

Answer is 3?

.solved