Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
