Question

If `f(9)=9` and `f^'(9)=4`, then what is the value of `lim_(x to9)(sqrt(f(x))-3)/(sqrtx -3)?`

Answer 1

`lim_(xto9) (sqrtf(x)-3)/(sqrtx-3)=4`

Explanation:

We know that,

`(1)f'(c)=lim_(xtoc) (f(x)-f(c))/(x-c) ,` take `x=9`

`=>f'(9)=lim_(xto9) (f(x)-f(9))/(x-9) ,where, f(9)=9`

`=>f'(9)=lim_(xto9)(f(x)-9)/(x-9) ,where, f'(9)=4`

`=>color(brown)(lim_(xto9)(f(x)-9)/(x-9)=4.....(2)`

We take,

`L=lim_(xto9) (sqrtf(x)-3)/(sqrtx-3)`
#color(white)(L)=lim_(xto9)(sqrtf(x)-3)/(sqrtx-3)xxcolor(red) ((1))xxcolor(blue)((1))#

`color(white)(L)=lim_(xto9)(sqrtf(x)-3)/(sqrtx-3)xxcolor(red)((sqrtf(x)+3)/(sqrtf(x)+3))xxcolor(blue)((sqrtx+3)/(sqrtx+3))`

`color(white)(L)=lim_(xto9)((sqrtf(x))^2-(3)^2)/((sqrtx)^2-(3)^2)xx(color(blue)(sqrtx+3))/(color(red)(sqrtf(x)+3))`

`color(white)(L)=color(brown)(lim_(xto9)(f(x)-9)/(x-9))xxlim_(xto9)(sqrtx+3)/(sqrtf(x)+3)...tocolor(brown)([use (2)]`

`color(white)(L)=color(brown)(4)xx(sqrt9+3)/(sqrtf(9)+3)`

`color(white)(L)=4[(3+3)/(sqrt9+3)]`

`:.L=4{6/6}=4`