Question

limit x tending to 0 (1/x-1/sinx)​

Answer 1

Answer:

The answer is 0

Step-by-step explanation:

Given limit is

     $=\displaystyle \lim_{x\to 0}(\frac{1}{x}-\frac{1}{\sin x})=\displaystyle \lim_{x\to 0}(\frac{\sin x-x}{x\sin x})$

$\boxed {\text{Using series expansion } \sin x=x-\frac{x^3}{6}+...}$

      $=\displaystyle \lim_{x\to 0}\frac{(x-\frac{x^3}{3}+..)-x}{x(x-\frac{x^3}{3}+...)}\\\\=\displaystyle \lim_{x\to 0}\frac{-\frac{x^3}{3}+..)}{x^2(1-\frac{x^2}{3}+...)}$

      $=\displaystyle \lim_{x\to 0}\frac{-\frac{x}{3}+..)}{(1-\frac{x^2}{3}+...)}$        $\boxed{\text{cancel }x^2 \text{ from numerator and denominator}}$

      $=\displaystyle \frac{-\frac{0}{3}+..)}{(1-\frac{0}{3}+...)} =0$