$0 < x < 1$ માટે,$\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ નું વિસ્તરણ શું છે?

• A
  $1+\frac{1}{2 x}-\frac{1}{2 !}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3 !}\left(\frac{1}{2 x}\right)^3-\frac{1 \cdot 3 \cdot 5}{4 !}\left(\frac{1}{2 x}\right)^4+\ldots \infty$
• B
  $\frac{1}{\sqrt{x}}+\frac{1}{2} \sqrt{x}-\frac{1}{2 !} \frac{x \sqrt{x}}{2^2}+\frac{1 \cdot 3}{3 !} \frac{x^2 \sqrt{x}}{2^3}-\ldots \infty$
• C
  $1+\frac{1}{\sqrt{x}}+\frac{1}{2} x \sqrt{x}+\frac{1}{2 !} \frac{x^2 \sqrt{x}}{2^3}+\frac{1 \cdot 3}{3 !} \frac{x^3 \sqrt{x}}{2^4}+\ldots \infty$
• D
  $\frac{1}{\sqrt{x}}+\frac{1}{2 x \sqrt{x}}-\frac{1}{2 !}\left(\frac{1}{2 x}\right)^2 \frac{1}{\sqrt{x}}+\frac{1 \cdot 3}{3 !}\left(\frac{1}{2 x}\right)^3 \frac{1}{\sqrt{x}}-\ldots \infty$