Question

lim x tends to pie/2 root2- root 1+sinx/root 2 cos^2x​

Answer 1

2 cos a is the correct answer

Answer 2

The solution of the given limit is

$\lim_{x \to \frac{\pi }{2} \1} \frac{\sqrt{2}-\sqrt{1+sinx} }{\sqrt{2cos^{2} x} } = 0$

• For solving the given limit,

  $\lim_{x \to \frac{\pi }{2} \1} \frac{\sqrt{2}-\sqrt{1+sinx} }{\sqrt{2cos^{2} x} }$

  $\lim_{x \to \frac{\pi }{2} \1} \frac{\sqrt{2}-\sqrt{1+sinx} }{\sqrt{2} cosx} }$       - (1)

• Take  $x = \frac{\pi }{2} - h$

     ∴         $h = \frac{\pi }{2} - x$

    Now as x→ $\frac{\pi }{2}$

             h→ 0

Now, putting the value of 'x' in (1) , the given limit becomes,

         $\lim_{h \to 0 \1} \frac{\sqrt{2}-\sqrt{1+sin(\frac{\pi }{2}-h) } }{\sqrt{2}cos(\frac{\pi }{2} -h)} }$

• Using the formulas

       $sin(\frac{\pi }{2} -h) = cosh\\and cos(\frac{\pi }{2}-h) = sinh$

       we get

          $\lim_{h \to 0 \1} \frac{\sqrt{2}-\sqrt{1+cosh} }{\sqrt{2}sinh} }$        - (2)

• Now, Using the identity

        $cos2x = 2cos^{2}x - 1 \\ 1 + cos2x = 2cos^{2} x$

        Here we have h = 2x and therefore x = $\frac{h}{2}$

        Now, (2) becomes

          $\lim_{h \to 0 \1} \frac{\sqrt{2}-\sqrt{2cos^{2} \frac{h}{2} } }{\sqrt{2}sinh} }$

          $\lim_{h \to 0 \1} \frac{\sqrt{2}-\sqrt{2}cos \frac{h}{2} }{\sqrt{2}sinh} }$

          $\lim_{h \to 0 \1} \frac{\sqrt{2}(1-cos \frac{h}{2} ) }{\sqrt{2}sinh} }$

          $\lim_{h \to 0 \1} \frac{(1-cos \frac{h}{2} ) }{sinh} }$

• Now, Using the identity

         $cos2x = 1 - 2sin^{2}x \\1-cos2x = 2sin^{2}x$

         we get,

         $\lim_{h \to 0 \1} \frac{2sin^{2}\frac{h}{4} }{sinh} }$

• dividing both numerator and denominator by 'h', we get

          $\lim_{h \to 0 \1} \ \frac{\frac{2sin^{2}\frac{h}{4} }{\frac{h^{2} }{16} \frac{16}{h} } }{\frac{sinh}{h} } }$

           $\frac{h}{8} \frac{ \lim_{h \to 0\1} \frac{sin^{2}\frac{h}{4} }{(\frac{h}{4}) ^{2} } }{ \lim_{h \to 0\1} \frac{sinh}{h} }$

• By using standard sine limit that is

          $\lim_{x \to 0\1} \frac{sinx}{x} = 1$, we get

          $\frac{h}{8} \frac{ \lim_{h \to 0\1} \frac{sin^{2}\frac{h}{4} }{(\frac{h}{4}) ^{2} } }{ \lim_{h \to 0\1} \frac{sinh}{h} } = \lim_{h \to 0\1} \frac{h}{8} (\frac{1}{1} )$

          $\lim_{h \to 0\1} \frac{h}{8} = 0$

• Therefore the required solution of the given limit is,

          $\lim_{x \to \frac{\pi }{2} \1} \frac{\sqrt{2}-\sqrt{1+sinx} }{\sqrt{2cos^{2} x} } = 0$