Question

$\underset{x\to 4}{lim}\frac{{\left(\cos \theta \right)}^x-{\left(\sin x\right)}^x-\cos 2\theta }{x-4}=$

• A. ${\cos }^4\theta \ln \cos \theta -{\sin }^4\theta \ln \sin \theta$
• B. ${\cos }^4\theta \ln \cos \theta +{\sin }^4\theta \ln \sin \theta$
• C. ${\cos }^4\theta \ln \sin \theta -{\sin }^4\theta \ln \cos \theta$
• D. None of these

Answer 1

The correct option is A ${\cos }^4\theta \ln \cos \theta -{\sin }^4\theta \ln \sin \theta$

$\underset{x\to 4}{lim}\frac{{\left(\cos \theta \right)}^x-{\left(\sin \theta \right)}^x-\cos 2\theta }{x-4}$

We can write $cos2\theta =co{s}^2\theta -si{n}^2\theta$

$cos2\theta =(co{s}^2\theta -si{n}^2\theta )(co{s}^2\theta +si{n}^2\theta )=co{s}^4\theta -si{n}^4\theta$

$=\underset{y\to 0}{lim}\frac{{\left(\cos \theta \right)}^{y+4}-{\left(\sin x\right)}^{y+4}-({\cos }^4\theta -{\sin }^4\theta )}{y}$ ....$[$Substituting $x-4=y$ and $\cos 2\theta ={\cos }^4\theta -{\sin }^4\theta ]$

$=\underset{y\to 0}{lim}{\cos }^4\theta \left[\frac{{\left(\cos \theta \right)}^y-1}{y}\right]-{\sin }^4\theta \left[\frac{{\left(\sin \theta \right)}^y-1}{y}\right]$

$={\cos }^4\theta \ln \cos \theta -{\sin }^4\theta \ln \sin \theta$ .......Using formula : $\underset{y\to 0}{lim}\frac{a^y-1}{y}=lna$