Question

find the limit.
a)
\lim_{t \to 0} \frac{\tan 6t}{\sin 2t}

Explanation:

Step1: Recall trigonometric limit rules

We know that \(\lim_{x
ightarrow0}\frac{\tan x}{x} = 1\) and \(\lim_{x
ightarrow0}\frac{\sin x}{x}=1\). Rewrite the given limit:
\[
\lim_{t
ightarrow0}\frac{\tan 6t}{\sin 2t}=\lim_{t
ightarrow0}\frac{\tan 6t}{6t}\cdot\frac{6t}{2t}\cdot\frac{2t}{\sin 2t}
\]

Step2: Apply limit properties

Using the product rule of limits \(\lim_{t
ightarrow a}(f(t)g(t)h(t))=\lim_{t
ightarrow a}f(t)\cdot\lim_{t
ightarrow a}g(t)\cdot\lim_{t
ightarrow a}h(t)\):

• \(\lim_{t

ightarrow0}\frac{\tan 6t}{6t} = 1\) (let \(x = 6t\), as \(t
ightarrow0\), \(x
ightarrow0\))

• \(\lim_{t

ightarrow0}\frac{6t}{2t}=\lim_{t
ightarrow0}3 = 3\)

• \(\lim_{t

ightarrow0}\frac{2t}{\sin 2t}=1\) (let \(x = 2t\), as \(t
ightarrow0\), \(x
ightarrow0\), and \(\lim_{x
ightarrow0}\frac{x}{\sin x}=1\))

Multiply these limits together: \(1\times3\times1 = 3\)

Answer:

\(3\)