Question

Evaluate $\underset{x\to 2}{lim}\frac{(\cos \alpha {)}^x+(\sin \alpha {)}^x-1}{x-2}$

• A. ${\cos }^2\alpha ln\cos \alpha +{\sin }^2\alpha ln\sin \alpha$
• B. ${\sin }^2\alpha ln\cos \alpha +{\cos }^2\alpha ln\sin \alpha$
• C. ${\sin }^2\alpha ln\cos \alpha -{\cos }^2\alpha ln\sin \alpha$
• D. ${\cos }^2\alpha ln\cos \alpha -{\sin }^2\alpha ln\sin \alpha$

Answer 1

The correct option is A ${\cos }^2\alpha ln\cos \alpha +{\sin }^2\alpha ln\sin \alpha$

$\underset{x\to 2}{lim}\frac{(\cos \alpha {)}^x+(\sin \alpha {)}^x-1}{x-2}$
$=\underset{h\to 0}{lim}\frac{{\cos }^2\alpha {\cos }^h\alpha +{\sin }^2\alpha {\sin }^h\alpha -1}{h}$
$=\underset{h\to 0}{lim}\frac{{\cos }^h\alpha -{\sin }^2\alpha {\cos }^h\alpha +{\sin }^2\alpha {\sin }^h\alpha -1}{h}$
$=\underset{h\to 0}{lim}\frac{{\cos }^h\alpha -1}{h}+{\sin }^2\alpha \left[\frac{{\sin }^h\alpha -1-{\cos }^h\alpha +1}{h}\right]$
$=\ell n\cos \alpha +{\sin }^2\alpha (\ell n\sin \alpha -\ell n\cos \alpha )$
$={\cos }^2\alpha \ell n\cos \alpha +{\sin }^2\alpha \ell n\sin \alpha$.