FORM GR1268 Question 1

Question:
\(\displaystyle\lim_{x\rightarrow0}\frac{\cos(3x)-1}{x^2}=\)

(A) \(\displaystyle\frac{9}{2}\)     (B) \(\displaystyle\frac{3}{2}\)     (C) \(\displaystyle-\frac{2}{3}\)     (D) \(\displaystyle-\frac{3}{2}\)     (E) \(\displaystyle-\frac{9}{2}\)

Answer:
(E)

Answer Key:
Since \(\cos(3\cdot0)-1=1-1=0\) and \(0^2=0\), by L'Hospital rule,
\[\lim_{x\rightarrow0}\frac{\cos(3x)-1}{x^2}
=\lim_{x\rightarrow0}\frac{(\cos(3x)-1)'}{(x^2)'}
=\lim_{x\rightarrow0}\frac{-3\sin(3x)}{2x}.\]

 Again, since \(-3\sin(3\cdot0)=0\) and \(2\cdot0=0\), by L'Hospital rule,
\[\begin{align}
\lim_{x\rightarrow0}\frac{-3\sin(3x)}{2x}
&=\lim_{x\rightarrow0}\frac{(-3\sin(3x))'}{(2x)'}\\
&=\lim_{x\rightarrow0}\frac{-9\cos(3x)}{2}\\
&=\frac{-9\cos(3\cdot0)}{2}\\
&=\frac{-9}{2}.
\end{align}\]