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.
