Question

Solve $\underset{x\to \pi /4}{lim}\frac{4\sqrt{2}-(\cos x+\sin x{)}^5}{1-\sin 2x}$

• A. $5$
• B. $5\sqrt{2}$
• C. $\sqrt{2}$
• D. None of these

Answer 1

The correct option is B $5\sqrt{2}$

The given limit.

$\underset{x\to \frac{\pi }{4}}{lim}\frac{4\sqrt{2}-{(cosx+sinx)}^5}{1-sin2x}$

By directly substituting limit value, we get

$\frac{0}{0}$

which is indeterminate value.

In order to find limit we apply L' Hospital's rule

Differentiate the numerator and denominator,

$\underset{x\to \frac{\pi }{4}}{lim}\frac{0-5.{(cosx+sinx)}^4.(-sinx+cosx)}{2.(sinx-cosx).(cosx+sinx)}$

Simplifying above,

$\underset{x\to \frac{\pi }{4}}{lim}\frac{5.{(cosx+sinx)}^4}{2.(cosx+sinx)}$

Or

$\underset{x\to \frac{\pi }{4}}{lim}\frac{5}{2}.{(cosx+sinx)}^3$

Now putting the limit and solving, we get

$=\frac{5}{2}.{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})}^3=\frac{5}{2}.{\left(\sqrt{2}\right)}^3=5\sqrt{2}$

which is the value.