Question

limit as x approaches 0 (x^2-sin^2(3x)) / 2x^2

Answer 1

lim (x-->0) sin²(3x)/x²
= lim (x-->0) [sin(3x)/x]²
= [lim (x-->0) sin(3x)/x]²
= [3 * lim (x-->0) sin(3x)/(3x)]²
= (3 * 1)²
= 9.

Answer 2

https://answers.yahoo.com/question/index?qid=20100202100353AAJyes2

Answer 3

no, the answer is -4, I just don't know how to find it. Can someone help me?

Answer 4

\[\begin{align*}\frac{x^2-\sin^23x}{2x^2}&=\frac{1}{2}-\frac{\sin^23x}{2x^2}\\
&=\frac{1}{2}-\frac{9\sin^23x}{2(9x^2)}\\
&=\frac{1}{2}-\frac{9}{2}\left(\frac{\sin 3x}{3x}\right)^2
\end{align*}\]
As \(x\to0\), the \(\dfrac{\sin3x}{3x}\) term approaches 1, leaving you with \(\dfrac{1}{2}-\dfrac{9}{2}=-4\).

Answer 5

thank you!!