(i) let L = $\displaystyle\lim_{x\to0} $$\frac{(sin x)sin^{-1}x - x^2}{x^6}$ Using standard series expansion for sin x and $sin^{-1}x$ L = $\displaystyle\lim_{x\to0}$$\dfrac{(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - ...)(x + \dfrac{x^3}{6} + \dfrac{3x^5}{40} ....) - x^2}{x^6}$ = $\displaystyle\lim_{x\to0}$$\dfrac{x^2 + \dfrac{x^4}{6} + \dfrac{3x^6}{40} - \dfrac{x^4}{6} - \dfrac{x^6}{36} + \dfrac{x^6}{120} + .... - x^2}{x^6}$ = $\displaystyle\lim_{x\to0}$$\dfrac{\dfrac{3x^6}{40} - \dfrac{x^6}{36} + \dfrac{x^6}{120}}{x^6}$ = $\displaystyle\lim_{x\to0}$$\dfrac{3}{40} - \dfrac{1}{36} + \dfrac{1}{120}$ = $\dfrac{20}{260}$ = $\dfrac{1}{18}$   (ii) let the line be y = a+ bx we have n = 5 | $x_i$ | $y_i$ | ${x_i}^2$ | ${y_i}^2$ | $x_iy_i$ | | --- | --- | --- | --- | --- | | -1 | -5 | 1 | 25 | 5 | | 1 | 1 | 1 | 1 | 1 | | 2 | 4 | 4 | 16 | 8 | | 3 | 7 | 9 | 49 | 21 | | 4 | 10 | 16 | 100 | 40 | we have $\sum{x_i}$ = 9 $\sum{y_i}$ = 17                             $\sum{{x_i}^2}$ = 31                               $\sum{x_iy_i}$ = 75 The equations are  $\sum y = na + b\sum x$ $\sum {xy} = a\sum x + b\sum {x^2}$

hence we have

17 = 5a + 9b              and                  75 = 9a + 31b

Solving the two equations simultaneously we get, a = -2 and b = 3

hence the equation of the line is y = -2 + 3x

at x = 7, y = -2 +3(7) = 19

thus y will be 19.