Let f(x)= x^1/2. Find f'(9) using the definition f'(a)=limex->a f(x)-f(a)/x-a

Question

I am stuck can you please help?

 

So far I have lim_x->9√x-√9

                              x  -9

 

                            =x(x^-1/2)-(3/x)

                              x(1-(9/x))

 

However the bottom still =0?

Answer 1

f'(9) = lim_x→9{(f(x)-f(9))/(x-9)}

 

       = lim_x→9{(√x - √9)/(x-9)}

 

       = lim_x→9{(√x - 3)/(x - 9)}  Plugging in x = 9 yields the                                                                indeterminate form 0/0

 

         Multiply the numerator and denominator by √x + 3 to get:

 

       = lim_x→9{(x-9)/[(x-9)(√x+3)]}

 

         ₌ lim_x→9{1/(√x+3)} = 1/(√9+3) = 1/6