Question

\[ \mathop {\lim }\limits_{x \to \infty } \frac{\mathrm{e}^{x^4}-1}{\mathrm{e}^{x^4}+1}= \]

• A. 1
• B. e
• C. \(\frac{1}{\mathrm{e}}\)
• D. not defined

Hint:

As x tends to infinity, \(e^{x^4}\) becomes very large, so the ratio approaches 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Divide the numerator and denominator by \(e^{x^4}\) to remove the dominant term.

Step 2: Simplify
\(\dfrac{e^{x^4}-1}{e^{x^4}+1}=\dfrac{1-e^{-x^4}}{1+e^{-x^4}}\)

Step 3: Take the Limit
As \(x\rightarrow\infty\), \(e^{-x^4}\rightarrow0\), so the limit is \(\dfrac{1-0}{1+0}=1\).

Final Answer:
The limit equals 1. \[ \boxed{\text{(A) }1} \]