How do you find the asymptotes for #y=x*ln[e + (1/x)] #?

Question

Elijah Allen · Accepted Answer

Asymptote at #x=-1/e,0# There is an asymptote at #x=0# because at this point the function is undefined as #1/0# is undefined. There is also an asymptote at #e+1/x=0# as the #ln# function is undefined at #0# #e+1/x=0 , ex+1=0 , ex=-1 x=-1/e#

Greyson Brown · Answer

undefined To find the asymptotes of the function $ y = x \cdot \ln\left(e + \frac{1}{x}\right) $, we need to examine the behavior of the function as $ x $ approaches certain values.

1. Vertical Asymptote:
   As $ x $ approaches zero from the positive side ($ x \rightarrow 0^+ $), the term $ \frac{1}{x} $ approaches positive infinity, and consequently, $ \ln(e + \frac{1}{x}) $ approaches positive infinity. Thus, there is a vertical asymptote at $ x = 0 $.

2. Horizontal Asymptote:
   As $ x $ approaches positive or negative infinity ($ x \rightarrow \pm \infty $), the term $ \frac{1}{x} $ approaches zero, and $ \ln(e + \frac{1}{x}) $ approaches $ \ln(e) = 1 $. Therefore, there is a horizontal asymptote at $ y = x \cdot 1 = x $.

Thus, the asymptotes for the function $ y = x \cdot \ln\left(e + \frac{1}{x}\right) $ are:

- Vertical asymptote: $ x = 0 $
- Horizontal asymptote: $ y = x $

HIX Tutor · Answer

undefined To find the asymptotes for the function $ y = x \ln\left(e + \frac{1}{x}\right) $, follow these steps:

1. **Identify Vertical Asymptotes**:
   - Look for values where the function is undefined. Here, $ \frac{1}{x} $ is undefined when $ x = 0 $. 
   - So, there is a **vertical asymptote at** $ x = 0 $.

2. **Identify Horizontal Asymptotes**:
   - Check the behavior of $ y $ as $ x $ goes to $ \infty $ and $ -\infty $.

- **As $ x \to \infty $**:
     - $ \frac{1}{x} $ becomes very small. So, $ e + \frac{1}{x} \approx e $.
     - Then, $ \ln(e + \frac{1}{x}) \approx \ln(e) = 1 $.
     - So, $ y \approx x \cdot 1 = x $. 
     - This means there's **no horizontal asymptote as $ y $ goes to $ \infty $**.

- **As $ x \to -\infty $**:
     - Again, $ \frac{1}{x} $ becomes very small (but negative). So, $ e + \frac{1}{x} \approx e $.
     - Similarly, $ \ln(e + \frac{1}{x}) \approx 1 $.
     - Then, $ y \approx x \cdot 1 = x $. 
     - This means there's **no horizontal asymptote as $ y $ goes to $ -\infty $**.

3. **Conclusion**:
   - There is a **vertical asymptote at** $ x = 0 $.
   - There are **no horizontal asymptotes**.

That's how you find the asymptotes!