Question

assignment 3: problem (1 point) evaluate $lim_{t
ightarrow0}\frac{sin 4t}{sin 9t}$. limit:

Explanation:

Step1: Use the limit - formula $\lim_{x

ightarrow0}\frac{\sin x}{x}=1$
We rewrite $\lim_{t
ightarrow0}\frac{\sin4t}{\sin9t}$ as $\lim_{t
ightarrow0}\frac{\sin4t}{4t}\cdot\frac{9t}{\sin9t}\cdot\frac{4}{9}$.

Step2: Apply the limit - formula

Since $\lim_{t
ightarrow0}\frac{\sin4t}{4t} = 1$ and $\lim_{t
ightarrow0}\frac{9t}{\sin9t}=1$, then $\lim_{t
ightarrow0}\frac{\sin4t}{4t}\cdot\frac{9t}{\sin9t}\cdot\frac{4}{9}=1\times1\times\frac{4}{9}$.

Answer:

$\frac{4}{9}$