લક્ષની કિંમત શોધો: $\mathop {\text{Limit}}\limits_{x \to 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}$,જ્યાં $0 < \alpha < \frac{\pi}{2}$.

  • A
    $(\cos \alpha)^4 \ln(\cos \alpha) + (\sin \alpha)^4 \ln(\sin \alpha)$
  • B
    $-(\cos \alpha)^4 \ln(\cos \alpha) - (\sin \alpha)^4 \ln(\sin \alpha)$
  • C
    $(\cos \alpha)^4 \ln(\cos \alpha) - (\sin \alpha)^4 \ln(\sin \alpha)$
  • D
    $-(\cos \alpha)^4 \ln(\cos \alpha) + (\sin \alpha)^4 \ln(\sin \alpha)$

Explore More

Similar Questions

$\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$

ધારો કે $x \neq 1$ માટે $g(x) = \frac{(x-1)^n}{\log \cos^m(x-1)}$ છે,અને ધારો કે $p$ એ $x=1$ આગળ $|x-1|$ નું ડાબી બાજુનું વિકલિત છે. જો $\lim_{x \rightarrow 1^{+}} g(x) = p$ હોય,તો:

ધારો કે $f(x)$ એ $x = h$ આગળ વિકલનીય છે. તો $\lim_{x \to h} \frac{(x + h)f(x) - 2hf(h)}{x - h}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {3x - a} - \sqrt {x + a} }}{{x - a}} = $

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x)\tan x$ ની કિંમત શોધો.

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