1explaination of q-06
3 Answers
13\frac{1}{x^2}-\frac{1}{x.Sin(x)} = \frac{Sin(x)-x}{x^2.Sin(x)}\\ \\ Use \ taylor \ series \ expansion\ , \frac{\left( x-\frac{x^3}{6}+...\right)-x}{x^2.\left(x-\frac{x^3}{6}+...\right)} \\ \\ \\ Which \ reduces \ to , \ \lim_{x \rightarrow 0} \frac{\left(\frac{-1}{6}+ x(...)\right)}{1- x(...)} = -1/6
