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Sagot :
To find the asymptotes of the curve [tex]\( y=\sqrt{1+x^2} \sin \frac{1}{x} \)[/tex], we need to analyze the behavior of the function as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] and [tex]\( -\infty \)[/tex].
### Step-by-Step Solution:
1. Identify the components of the function:
- The function is composed of two parts: [tex]\( \sqrt{1+x^2} \)[/tex] and [tex]\( \sin \frac{1}{x} \)[/tex].
2. Examine the behavior as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex]:
- Consider the term [tex]\( \sqrt{1+x^2} \)[/tex]:
[tex]\[ \sqrt{1+x^2} \approx x \text{ as } x \to \infty \text{ and } x \to -\infty \][/tex]
- Now, consider the behavior of [tex]\( \sin \frac{1}{x} \)[/tex]:
[tex]\[ \sin \frac{1}{x} \text{ oscillates between } -1 \text{ and } 1 \text{ as } x \to \infty \text{ and } x \to -\infty \][/tex]
3. Combine the behaviors:
- As [tex]\( x \to \infty \)[/tex], the product [tex]\( \sqrt{1+x^2} \sin \frac{1}{x} \)[/tex] would oscillate between [tex]\( \sqrt{1+x^2} \cdot (-1) \)[/tex] and [tex]\( \sqrt{1+x^2} \cdot (+1) \)[/tex], which simplifies to:
[tex]\[ -\sqrt{1+x^2} \leq \sqrt{1+x^2} \sin \frac{1}{x} \leq \sqrt{1+x^2} \][/tex]
- Since [tex]\( \sqrt{1+x^2} \approx x \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex], we have:
[tex]\[ -x \leq \sqrt{1+x^2} \sin \frac{1}{x} \leq x \][/tex]
4. Calculate the limits to confirm the end behavior:
- The limit of [tex]\( y \)[/tex] as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} \sqrt{1+x^2} \sin \frac{1}{x} = 1 \][/tex]
- The limit of [tex]\( y \)[/tex] as [tex]\( x \to -\infty \)[/tex]:
[tex]\[ \lim_{x \to -\infty} \sqrt{1+x^2} \sin \frac{1}{x} = -1 \][/tex]
### Conclusion:
There are no vertical or horizontal asymptotes in the classical sense, but the function approaches the lines [tex]\( y = 1 \)[/tex] and [tex]\( y = -1 \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex], respectively. Therefore, these lines can be interpreted as the horizontal boundary-like asymptotes for the behavior of the given function at infinity.
### Step-by-Step Solution:
1. Identify the components of the function:
- The function is composed of two parts: [tex]\( \sqrt{1+x^2} \)[/tex] and [tex]\( \sin \frac{1}{x} \)[/tex].
2. Examine the behavior as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex]:
- Consider the term [tex]\( \sqrt{1+x^2} \)[/tex]:
[tex]\[ \sqrt{1+x^2} \approx x \text{ as } x \to \infty \text{ and } x \to -\infty \][/tex]
- Now, consider the behavior of [tex]\( \sin \frac{1}{x} \)[/tex]:
[tex]\[ \sin \frac{1}{x} \text{ oscillates between } -1 \text{ and } 1 \text{ as } x \to \infty \text{ and } x \to -\infty \][/tex]
3. Combine the behaviors:
- As [tex]\( x \to \infty \)[/tex], the product [tex]\( \sqrt{1+x^2} \sin \frac{1}{x} \)[/tex] would oscillate between [tex]\( \sqrt{1+x^2} \cdot (-1) \)[/tex] and [tex]\( \sqrt{1+x^2} \cdot (+1) \)[/tex], which simplifies to:
[tex]\[ -\sqrt{1+x^2} \leq \sqrt{1+x^2} \sin \frac{1}{x} \leq \sqrt{1+x^2} \][/tex]
- Since [tex]\( \sqrt{1+x^2} \approx x \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex], we have:
[tex]\[ -x \leq \sqrt{1+x^2} \sin \frac{1}{x} \leq x \][/tex]
4. Calculate the limits to confirm the end behavior:
- The limit of [tex]\( y \)[/tex] as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} \sqrt{1+x^2} \sin \frac{1}{x} = 1 \][/tex]
- The limit of [tex]\( y \)[/tex] as [tex]\( x \to -\infty \)[/tex]:
[tex]\[ \lim_{x \to -\infty} \sqrt{1+x^2} \sin \frac{1}{x} = -1 \][/tex]
### Conclusion:
There are no vertical or horizontal asymptotes in the classical sense, but the function approaches the lines [tex]\( y = 1 \)[/tex] and [tex]\( y = -1 \)[/tex] as [tex]\( x \to \infty \)[/tex] and [tex]\( x \to -\infty \)[/tex], respectively. Therefore, these lines can be interpreted as the horizontal boundary-like asymptotes for the behavior of the given function at infinity.
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