Question

Calculus: Suppose that f'(2)=3. Find limit as x approaches 2 of [f(x)-f(2)]/[sqrt(x)-sqrt(2)]??

Answer 1

`lim_(x->2) (f(x)-f(2))/(sqrtx-sqrt2)=6sqrt2`

Explanation:

We know that,

`(1)f'(x)=lim_(x->c) (f(x)-f(c))/(x-c)`

Using `(1)`we get

`f'(2)=3`

`f'(2)=color(red)(lim_(x->2) (f(x)-f(2))/(x-2)=3...to(A)`

So, we take

`L=lim_(x->2) (f(x)-f(2))/(sqrtx-sqrt2)`

#L=lim_(x->2) (f(x)-f(2))/(sqrtx- sqrt2)xx(sqrtx+sqrt2)/(sqrtx+sqrt2)#

`L=color(red)(lim_(x->2) (f(x)-f(2))/(x-2))xxlim_(x->2)(sqrtx+sqrt2)`

`L=color(red)(3)xx(sqrt2+sqrt2)...tocolor(red)(Apply(A)`

`L=3xx2sqrt2`

`L=6sqrt2`