Question

$\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}-(cosx+sinx{)}^3}{1-sin2x}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}\{1-\frac{1}{2\sqrt{2}}(cosx+sinx{)}^3\}}{(1-sin2x)}\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}\{1-{(\frac{1}{\sqrt{2}}cosx+\frac{1}{\sqrt{2}}sinx)}^3\}}{(1-sin2x)}=\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}\{1-{(\frac{1}{\sqrt{2}}cosx+\frac{1}{\sqrt{2}}sinx)}^3\}}{(1-sin2x)}\underset{x\to \frac{\pi }{4}}{lim}\frac{2\sqrt{2}\{1-co{s}^3(\frac{\pi }{4}-x)\}}{(1-sin2x)}Putx=\frac{\pi }{4}+h,then=\underset{h\to 0}{lim}\frac{2\sqrt{2}(1-co{s}^{3}h)}{(1-cos2h)}=\underset{h\to 0}{lim}\frac{2\sqrt{2}(1-co{s}^{3}h)}{\left(2si{n}^{2}h\right)}=\underset{h\to 0}{lim}\frac{2\sqrt{2}(1-cosh)(1+cosh+co{s}^{2}h)}{2(1-cosh)(1+cosh)}=\underset{h\to 0}{lim}\frac{\sqrt{2}(1+cosh+co{s}^{2}h)}{(1+cosh)}=\frac{3\sqrt{2}}{2}$