Help finding the slope of a tangent line using limit?

Question

Given f(x)= 1/(x+3), find the slope of a tangent line at (2,1/5) using a limit.

Please do not show me the derivative. I know it's faster, but we're supposed to learn how to use it using the f(a+h)- f(a)/ h formula.

I've tried it but keep getting stuck with messy fractions and I can't figure out how to isolate the h so I can cancel them on the top and bottom.

Help!!!

Answer 1

f(x) = 1 / (x + 3)

f(x + h) = 1 / (x + h + 3)

f(x + h) - f(x) = [ (x + 3) - (x + h + 3) ] / [ (x + 3)(x + h + 3) ]

= - h / [ (x + 3)(x + h + 3) ].

[ f(x + h) - f(x) ] / h = - 1 / [ (x + 3)(x + h + 3) ]

f'(x) = lim(x -> 0) [ f(x + h) - f(x) ] / h ]

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

f'(2) = - 1 / (2 + 3)^2

= - 1 / 5^2

= - 1 / 25.