$$ \lim _ { x \rightarrow - \infty } \frac { 1 + \sin ^ { 3 } x } { \cos ^ { 2 } x } $$

Question

UpStudy AI · Accepted Answer

To evaluate the limit:

$$
\lim_{x \to -\infty} \frac{1 + \sin^3 x}{\cos^2 x},
$$

let's analyze the behavior of the numerator and denominator as $$ x $$ approaches negative infinity.

### Behavior of the Functions

1. **Sine and Cosine Functions**:
   - Both $$ \sin x $$ and $$ \cos x $$ are **periodic** functions with period $$ 2\pi $$. As $$ x $$ approaches $$ -\infty $$, these functions **do not settle down** to any particular value but continue to oscillate between $$-1$$ and $$1$$ for $$ \sin x $$, and between $$-1$$ and $$1$$ for $$ \cos x $$.

2. **Numerator $$ 1 + \sin^3 x $$**:
   - Since $$ \sin x \in [-1, 1] $$, $$ \sin^3 x \in [-1, 1] $$.
   - Therefore, $$ 1 + \sin^3 x \in [0, 2] $$.

3. **Denominator $$ \cos^2 x $$**:
   - Since $$ \cos^2 x \geq 0 $$ for all $$ x $$, and $$ \cos^2 x \in [0, 1] $$.

### Analyzing the Limit

Given the oscillatory nature of both the numerator and the denominator:

- **When $$ \cos^2 x $$ is near its minimum (i.e., $$ \cos x \approx 0 $$)**:
  - The denominator approaches $$ 0 $$, causing the fraction $$ \frac{1 + \sin^3 x}{\cos^2 x} $$ to **grow without bound**, especially if $$ 1 + \sin^3 x $$ is not simultaneously approaching $$ 0 $$.

- **When $$ \cos^2 x $$ is near its maximum (i.e., $$ \cos x \approx \pm1 $$)**:
  - The denominator is around $$ 1 $$, and the numerator oscillates between $$ 0 $$ and $$ 2 $$, making the overall fraction oscillate between $$ 0 $$ and $$ 2 $$.

### Conclusion

Due to the periodic and unbounded behavior of the trigonometric functions involved:

- **The limit does not settle to a single finite value**.
- **The expression oscillates** between finite values and grows without bound as $$ x $$ approaches $$ -\infty $$.

Therefore, the limit **does not exist** in the conventional sense.

**Final Answer:**

The limit does not exist because the expression grows without bound as x approaches –∞.
The limit does not exist because the expression grows without bound as $$ x $$ approaches negative infinity.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

preguntas relacionadas

Latest Calculus Questions