Question

Find the limit as x approaches 0:
2sin^3 3x/-x^3
Any help would be appreciated! Thank you!

Answer 1

Hint : \[ \dfrac{\sin^3(3x)}{x^3} = 27 \cdot \dfrac{\sin^3(3x)}{(3x)^3}\]

Answer 2

\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\]

Answer 3

does it ring a bell

Answer 4

so can we say that limit of sin^3 3x / / 3x^3 has a limit of 1 as x approaches 0?

Answer 5

if so then the required limit is -2 8 27 = -54

Answer 6

* -2 * 27 = -54

Answer 7

i think this is valid - limits are not my strong point

Answer 8

So I'm going with -54 on that question. Could you also check my answer? I'm supposed to find the derivative of
f(x)= ((3x^2)/2) -x
I got 3x-1 for my answer

Answer 9

\(\LARGE \color{Red}{\checkmark}\)

Answer 10

Thanks to everyone so much!