Question

use lhôpitals rule to evaluate $lim_{x \to 0}\frac{3 - 3cos x}{8x^{2}}$. then determine the limit using limit laws and commonly known limits. use lhôpitals rule to rewrite the given limit so that it is not an indeterminate form. $lim_{x \to 0}\frac{3 - 3cos x}{8x^{2}}=lim_{x \to 0}(square)$

Explanation:

Step1: Check indeterminate form

When $x
ightarrow0$, $\lim_{x
ightarrow0}(3 - 3\cos x)=3-3\cos(0)=3 - 3=0$ and $\lim_{x
ightarrow0}(8x^{2})=0$. So it's in $\frac{0}{0}$ form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $3 - 3\cos x$ with respect to $x$ is $3\sin x$, and the derivative of $8x^{2}$ with respect to $x$ is $16x$. So $\lim_{x
ightarrow0}\frac{3 - 3\cos x}{8x^{2}}=\lim_{x
ightarrow0}\frac{3\sin x}{16x}$.

Step3: Check new form and apply L'Hopital's Rule again (if needed)

When $x
ightarrow0$, $\lim_{x
ightarrow0}(3\sin x)=0$ and $\lim_{x
ightarrow0}(16x)=0$. Apply L'Hopital's Rule again. The derivative of $3\sin x$ with respect to $x$ is $3\cos x$, and the derivative of $16x$ with respect to $x$ is $16$. So $\lim_{x
ightarrow0}\frac{3\sin x}{16x}=\lim_{x
ightarrow0}\frac{3\cos x}{16}$.

Step4: Evaluate the limit

Substitute $x = 0$ into $\frac{3\cos x}{16}$. We get $\frac{3\cos(0)}{16}=\frac{3\times1}{16}=\frac{3}{16}$.

Answer:

$\frac{3}{16}$