The graph of which of the following equations has y = 1 as an asymptote?A. y = ln(x)B.𝑦=𝑥𝑥+1y= x+1x​ C.𝑦=𝑒−𝑥y=e −x D.𝑦=𝑥2𝑥−1y= x−1x 2 ​ E. y = sin(x)

To determine which equation has $y = 1$ as an asymptote, we will analyze each option.

1. Option A: $y = \ln(x)$

   • The natural logarithm function does not approach 1 as $x$ increases. It approaches $\infty$ as $x$ goes to $\infty$.
   • Conclusion: No asymptote at $y = 1$.

2. Option B: $y = \frac{x + 1}{x}$

   • This can be rewritten as $y = 1 + \frac{1}{x}$. As $x$ approaches $\infty$, $\frac{1}{x}$ approaches 0. Thus, $y$ approaches 1.
   • Conclusion: This has a horizontal asymptote at $y = 1$.

3. Option C: $y = e^{-x}$

   • As $x$ goes to $\infty$, $e^{-x}$ approaches 0, not 1.
   • Conclusion: No asymptote at $y = 1$.

4. Option D: $y = \frac{x}{x^2 - 1}$

   • This function does not have a horizontal asymptote at 1. The horizontal asymptote can be found by analyzing $\lim_{x \to \infty} \frac{x}{x^2 - 1}$, which approaches 0.
   • Conclusion: No asymptote at $y = 1$.

5. Option E: $y = \sin(x)$

   • The sine function oscillates between -1 and 1. It does not have a horizontal asymptote.
   • Conclusion: No asymptote at $y = 1$.

Final Answer

The correct option is B. $y = \frac{x + 1}{x}$, as it approaches $y = 1$ as $x$ goes to infinity.