Question

question 5 · 1 point evaluate the following limit using lhospitals rule. enter an exact answer. provide your answer below: $lim_{x \to 0}\frac{sin(8x)}{16x}=$

Explanation:

Step1: Check form of limit

As $x
ightarrow0$, $\sin(8x)
ightarrow0$ and $16x
ightarrow0$. So, it's in $\frac{0}{0}$ form, and L'Hospital's rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of $\sin(8x)$ with respect to $x$ is $8\cos(8x)$ (by chain - rule, derivative of $\sin(u)$ is $\cos(u)$ and $u = 8x$, so $\frac{d}{dx}\sin(8x)=\cos(8x)\cdot8$), and the derivative of $16x$ with respect to $x$ is 16. So, $\lim_{x
ightarrow0}\frac{\sin(8x)}{16x}=\lim_{x
ightarrow0}\frac{8\cos(8x)}{16}$.

Step3: Evaluate the new limit

Substitute $x = 0$ into $\frac{8\cos(8x)}{16}$. Since $\cos(0)=1$, we have $\frac{8\cos(0)}{16}=\frac{8\times1}{16}=\frac{1}{2}$.

Answer:

$\frac{1}{2}$