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Sagot :
To find the asymptotes of the curve [tex]\(y = \sqrt{1 + x^2} \sin \left(\frac{1}{x} \right)\)[/tex], we need to examine both the horizontal and vertical asymptotes.
### Horizontal Asymptotes:
Horizontal asymptotes can be found by taking the limit of the function as [tex]\(x\)[/tex] approaches positive and negative infinity.
1. Limit as [tex]\(x\)[/tex] approaches [tex]\(\infty\)[/tex]:
[tex]\[ \lim_{x \to \infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) \][/tex]
As [tex]\(x\)[/tex] becomes very large, [tex]\(\frac{1}{x}\)[/tex] approaches 0. The sine function, [tex]\(\sin \left( \frac{1}{x} \right)\)[/tex], oscillates between -1 and 1, and since [tex]\(\sqrt{1 + x^2}\)[/tex] grows unbounded, it affects the overall limit. However, the important aspect here is to observe the behavior of the function normed on these oscillations as [tex]\(x\)[/tex] grows indefinitely.
Surprisingly, it can be seen that:
[tex]\[ \lim_{x \to \infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) = 1 \][/tex]
2. Limit as [tex]\(x\)[/tex] approaches [tex]\(-\infty\)[/tex]:
[tex]\[ \lim_{x \to -\infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) \][/tex]
Similarly, as [tex]\(x\)[/tex] becomes very large negative, [tex]\(\frac{1}{x}\)[/tex] approaches 0 from the negative side. The function [tex]\(\sin \left( \frac{1}{x} \right)\)[/tex] will still oscillate between -1 and 1. When multiplied by [tex]\(\sqrt{1 + x^2}\)[/tex], we focus on the behavior of multiplication affecting the sine wave amplitude normed to large values. Hence, the significant asymptotic value remains:
[tex]\[ \lim_{x \to -\infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) = -1 \][/tex]
Thus, the horizontal asymptotes are:
- As [tex]\( x \to \infty \)[/tex], [tex]\( y = 1 \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( y = -1 \)[/tex]
### Vertical Asymptotes:
Next, we need to determine if there are any vertical asymptotes, where the function potentially becomes unbounded as [tex]\(x\)[/tex] approaches some finite value.
3. For a function to have a vertical asymptote, it typically involves the denominator of a rational function approaching zero, or the function itself going to [tex]\(\pm\infty\)[/tex] at some finite [tex]\(x\)[/tex]. In our case, [tex]\(\sqrt{1 + x^2}\)[/tex] never equals zero as it’s always at least 1 (since [tex]\(1 + x^2 \geq 1\)[/tex] for all [tex]\(x \in \mathbb{R}\)[/tex]) and [tex]\(\sin \left( \frac{1}{x} \right)\)[/tex] is bounded between -1 and 1. Therefore, there are no points where [tex]\(y\)[/tex] approaches [tex]\(\pm\infty\)[/tex] for any finite [tex]\(x\)[/tex], indicating there are no vertical asymptotes.
Thus, the vertical asymptotes are:
- None.
So, the final answer is:
- Horizontal asymptotes: [tex]\(y = 1\)[/tex] as [tex]\(x \to \infty\)[/tex] and [tex]\(y = -1\)[/tex] as [tex]\(x \to -\infty\)[/tex]
- Vertical asymptotes: None.
### Horizontal Asymptotes:
Horizontal asymptotes can be found by taking the limit of the function as [tex]\(x\)[/tex] approaches positive and negative infinity.
1. Limit as [tex]\(x\)[/tex] approaches [tex]\(\infty\)[/tex]:
[tex]\[ \lim_{x \to \infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) \][/tex]
As [tex]\(x\)[/tex] becomes very large, [tex]\(\frac{1}{x}\)[/tex] approaches 0. The sine function, [tex]\(\sin \left( \frac{1}{x} \right)\)[/tex], oscillates between -1 and 1, and since [tex]\(\sqrt{1 + x^2}\)[/tex] grows unbounded, it affects the overall limit. However, the important aspect here is to observe the behavior of the function normed on these oscillations as [tex]\(x\)[/tex] grows indefinitely.
Surprisingly, it can be seen that:
[tex]\[ \lim_{x \to \infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) = 1 \][/tex]
2. Limit as [tex]\(x\)[/tex] approaches [tex]\(-\infty\)[/tex]:
[tex]\[ \lim_{x \to -\infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) \][/tex]
Similarly, as [tex]\(x\)[/tex] becomes very large negative, [tex]\(\frac{1}{x}\)[/tex] approaches 0 from the negative side. The function [tex]\(\sin \left( \frac{1}{x} \right)\)[/tex] will still oscillate between -1 and 1. When multiplied by [tex]\(\sqrt{1 + x^2}\)[/tex], we focus on the behavior of multiplication affecting the sine wave amplitude normed to large values. Hence, the significant asymptotic value remains:
[tex]\[ \lim_{x \to -\infty} \sqrt{1 + x^2} \sin \left( \frac{1}{x} \right) = -1 \][/tex]
Thus, the horizontal asymptotes are:
- As [tex]\( x \to \infty \)[/tex], [tex]\( y = 1 \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( y = -1 \)[/tex]
### Vertical Asymptotes:
Next, we need to determine if there are any vertical asymptotes, where the function potentially becomes unbounded as [tex]\(x\)[/tex] approaches some finite value.
3. For a function to have a vertical asymptote, it typically involves the denominator of a rational function approaching zero, or the function itself going to [tex]\(\pm\infty\)[/tex] at some finite [tex]\(x\)[/tex]. In our case, [tex]\(\sqrt{1 + x^2}\)[/tex] never equals zero as it’s always at least 1 (since [tex]\(1 + x^2 \geq 1\)[/tex] for all [tex]\(x \in \mathbb{R}\)[/tex]) and [tex]\(\sin \left( \frac{1}{x} \right)\)[/tex] is bounded between -1 and 1. Therefore, there are no points where [tex]\(y\)[/tex] approaches [tex]\(\pm\infty\)[/tex] for any finite [tex]\(x\)[/tex], indicating there are no vertical asymptotes.
Thus, the vertical asymptotes are:
- None.
So, the final answer is:
- Horizontal asymptotes: [tex]\(y = 1\)[/tex] as [tex]\(x \to \infty\)[/tex] and [tex]\(y = -1\)[/tex] as [tex]\(x \to -\infty\)[/tex]
- Vertical asymptotes: None.
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