Question

find the limit.
lim_{x
ightarrow0}\frac{sin(8x)}{9x^{3}-5x}

Explanation:

Step1: Factor out x from the denominator

We have $\lim_{x
ightarrow0}\frac{\sin(8x)}{x(9x^{2}-5)}$.

Step2: Use the limit - property $\lim_{u

ightarrow0}\frac{\sin(u)}{u}=1$
Let $u = 8x$. As $x
ightarrow0$, $u
ightarrow0$. So $\lim_{x
ightarrow0}\frac{\sin(8x)}{x}=\lim_{x
ightarrow0}\frac{\sin(8x)}{x}\cdot\frac{8}{8}=8\lim_{x
ightarrow0}\frac{\sin(8x)}{8x}=8$.

Step3: Evaluate the limit of the remaining part

$\lim_{x
ightarrow0}(9x^{2}-5)=9(0)^{2}-5=-5$.

Step4: Calculate the original limit

$\lim_{x
ightarrow0}\frac{\sin(8x)}{x(9x^{2}-5)}=\lim_{x
ightarrow0}\frac{\sin(8x)}{x}\cdot\lim_{x
ightarrow0}\frac{1}{9x^{2}-5}=8\times\frac{1}{-5}=-\frac{8}{5}$.

Answer:

$-\frac{8}{5}$