To find the limit of the function as [tex]\( t \)[/tex] approaches 0, we need to analyze the expression [tex]\(\sec\left(\frac{-2 \pi t}{\sin t}\right)\)[/tex].
1. Identify the function:
[tex]\[
\sec\left(\frac{-2 \pi t}{\sin t}\right)
\][/tex]
2. Understand the behavior of the trigonometric functions near [tex]\( t = 0 \)[/tex]:
- The function [tex]\(\sin(t)\)[/tex] has a known limit property:
[tex]\[
\lim_{t \to 0} \frac{\sin t}{t} = 1 \implies \sin(t) \approx t \quad \text{near } t = 0
\][/tex]
- Therefore, as [tex]\( t \)[/tex] approaches 0, [tex]\(\sin(t) \)[/tex] behaves very similarly to [tex]\( t \)[/tex]. Thus we can approximate:
[tex]\[
\frac{\sin t}{t} \approx 1
\][/tex]
3. Simplify the argument of the secant function:
- Substitute [tex]\(\sin t \approx t\)[/tex] in the argument:
[tex]\[
\frac{-2 \pi t}{\sin t} \approx \frac{-2 \pi t}{t} = -2\pi
\][/tex]
4. Substitute this approximation:
- As [tex]\( t \to 0 \)[/tex], [tex]\(\frac{-2 \pi t}{\sin t} \to -2\pi \)[/tex].
5. Evaluate the secant function at this limit:
- Recall that [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex]:
[tex]\[
\sec(-2\pi) = \frac{1}{\cos(-2\pi)}
\][/tex]
- The cosine function is periodic with period [tex]\( 2\pi \)[/tex], so:
[tex]\[
\cos(-2\pi) = \cos(2\pi) = \cos(0) = 1
\][/tex]
6. Compute the final value:
[tex]\[
\sec(-2\pi) = \frac{1}{\cos(-2\pi)} = \frac{1}{1} = 1
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{t \rightarrow 0} \sec\left(\frac{-2\pi t}{\sin t}\right) = 1
\][/tex]
The answer is [tex]\( \boxed{1} \)[/tex].
