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

  • A
    $1$
  • B
    $e$
  • C
    $\frac{1}{e}$
  • D
    $\text{not defined}$

Explore More

Similar Questions

$\mathop {\lim }\limits_{x \to a} f(x) \cdot g(x)$ का अस्तित्व है,यदि

यदि $a, b, c$ और $k$ शून्येतर वास्तविक संख्याएँ हैं और $\lim _{x \rightarrow \infty} x\left(a^{\frac{1}{x}}+b^{\frac{1}{x}}+c^{\frac{1}{x}}-3 k^{\frac{1}{x}}\right)=0$,तो $k=$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}=$

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\alpha x}} - {e^{\beta x}}}}{x} = $

$\mathop {\lim }\limits_{x \to 0} f(x)$ का मान ज्ञात कीजिए,जहाँ $f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo