Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To determine the asymptotes of the curve [tex]\( y = \sqrt{1 + x^2} \sin \frac{1}{x} \)[/tex], we need to examine the behavior of the function as [tex]\(x\)[/tex] approaches different critical values, particularly as [tex]\( x \to \infty \)[/tex], [tex]\( x \to -\infty \)[/tex], and [tex]\( x \to 0 \)[/tex].
### 1. Horizontal Asymptotes:
Horizontal asymptotes occur if the function approaches a specific finite value as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] or [tex]\( -\infty \)[/tex].
- As [tex]\( x \to \infty \)[/tex]:
The term [tex]\( \sqrt{1 + x^2} \)[/tex] grows without bound because [tex]\(1 + x^2\)[/tex] becomes very large and taking the square root of a very large number still results in a large number. On the other hand, [tex]\(\sin \frac{1}{x} \)[/tex] oscillates between -1 and 1, but does not settle to any particular value. As a result, the product [tex]\(\sqrt{1 + x^2} \sin \frac{1}{x}\)[/tex] also oscillates and does not approach a finite limit.
- As [tex]\( x \to -\infty \)[/tex]:
Similarly, the behavior when [tex]\( x \to -\infty \)[/tex] is analogous to [tex]\( x \to \infty \)[/tex]. Here, [tex]\(\sqrt{1 + x^2}\)[/tex] still grows large since the square root of a large positive number is always positive and large. [tex]\(\sin \frac{1}{x} \)[/tex] continues to oscillate between -1 and 1. Therefore, the product [tex]\(\sqrt{1 + x^2} \sin \frac{1}{x}\)[/tex] oscillates as well and does not approach a finite limit.
Thus, the function does not have horizontal asymptotes.
### 2. Vertical Asymptotes:
Vertical asymptotes commonly occur if the function heads to [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex] as [tex]\(x\)[/tex] approaches a certain finite value.
- As [tex]\( x \to 0 \)[/tex]:
Consider the limit as [tex]\( x \)[/tex] approaches zero. The term [tex]\(\frac{1}{x}\)[/tex] grows extremely large in magnitude, causing the sine function to oscillate very rapidly between -1 and 1. This rapid oscillation means [tex]\(\sin \frac{1}{x}\)[/tex] does not settle to any finite value, but rather, causes [tex]\( \sqrt{1 + x^2} \sin \frac{1}{x} \)[/tex] to oscillate uncontrollably. Therefore, there is no specific direction towards infinity or negative infinity.
Hence, there are no vertical asymptotes either.
### Conclusion:
After analyzing the function [tex]\( y = \sqrt{1 + x^2} \sin \frac{1}{x} \)[/tex] as [tex]\( x \)[/tex] approaches positive and negative infinity, as well as zero, we can conclude:
The curve [tex]\(y = \sqrt{1 + x^2} \sin \frac{1}{x} \)[/tex] does not have vertical or horizontal asymptotes.
### 1. Horizontal Asymptotes:
Horizontal asymptotes occur if the function approaches a specific finite value as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] or [tex]\( -\infty \)[/tex].
- As [tex]\( x \to \infty \)[/tex]:
The term [tex]\( \sqrt{1 + x^2} \)[/tex] grows without bound because [tex]\(1 + x^2\)[/tex] becomes very large and taking the square root of a very large number still results in a large number. On the other hand, [tex]\(\sin \frac{1}{x} \)[/tex] oscillates between -1 and 1, but does not settle to any particular value. As a result, the product [tex]\(\sqrt{1 + x^2} \sin \frac{1}{x}\)[/tex] also oscillates and does not approach a finite limit.
- As [tex]\( x \to -\infty \)[/tex]:
Similarly, the behavior when [tex]\( x \to -\infty \)[/tex] is analogous to [tex]\( x \to \infty \)[/tex]. Here, [tex]\(\sqrt{1 + x^2}\)[/tex] still grows large since the square root of a large positive number is always positive and large. [tex]\(\sin \frac{1}{x} \)[/tex] continues to oscillate between -1 and 1. Therefore, the product [tex]\(\sqrt{1 + x^2} \sin \frac{1}{x}\)[/tex] oscillates as well and does not approach a finite limit.
Thus, the function does not have horizontal asymptotes.
### 2. Vertical Asymptotes:
Vertical asymptotes commonly occur if the function heads to [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex] as [tex]\(x\)[/tex] approaches a certain finite value.
- As [tex]\( x \to 0 \)[/tex]:
Consider the limit as [tex]\( x \)[/tex] approaches zero. The term [tex]\(\frac{1}{x}\)[/tex] grows extremely large in magnitude, causing the sine function to oscillate very rapidly between -1 and 1. This rapid oscillation means [tex]\(\sin \frac{1}{x}\)[/tex] does not settle to any finite value, but rather, causes [tex]\( \sqrt{1 + x^2} \sin \frac{1}{x} \)[/tex] to oscillate uncontrollably. Therefore, there is no specific direction towards infinity or negative infinity.
Hence, there are no vertical asymptotes either.
### Conclusion:
After analyzing the function [tex]\( y = \sqrt{1 + x^2} \sin \frac{1}{x} \)[/tex] as [tex]\( x \)[/tex] approaches positive and negative infinity, as well as zero, we can conclude:
The curve [tex]\(y = \sqrt{1 + x^2} \sin \frac{1}{x} \)[/tex] does not have vertical or horizontal asymptotes.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
