Q53 / 150 · MHT CET 2025 (21 Apr Shift 1) · Mathematics

$\lim _{x \rightarrow \infty} \frac{\mathrm{e}^{x^4}-1}{\mathrm{e}^{x^4}+1}=$

  • Correct answer
    A
    $1$
  • B
    e
  • C
    $\frac{1}{\mathrm{e}}$
  • D
    not defined

Solution

Given the limit $\lim_{x \to \infty} \frac{e^{x^4} - 1}{e^{x^4} + 1}$, observe that as $x \to \infty$, $x^4 \to \infty$ and consequently $e^{x^4} \to \infty$.

Dividing numerator and denominator by $e^{x^4}$ yields $\frac{1 - e^{-x^4}}{1 + e^{-x^4}}$.

Since $-x^4 \to -\infty$ as $x \to \infty$, we have $e^{-x^4} \to 0$.

Substituting gives $\frac{1 - 0}{1 + 0} = 1$.

Hence, the correct answer is option: 'A'.

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