Calculus Help! definition of a derivative

Find the equation of the tangent line to f(x) = sqrt x at x=81 using the definition of a derivative.

Hint: f'(x)= lim h approaching 0 [f(x+h) - f(x)]/x

By definition,

f '(81) = lim [ ( √(81+h) - √81 ) / h ], as h -> 0

f '(81) = lim [ ( √(81+h) - 9 ) / h ], as h -> 0

The trick here is to rationalize the numerator by multiplying by √(81+h)+9 / √(81+h)+9

f '(81) = lim [ ( (81+h) - 81 ) / ( h (√81+h) +9 ) ], as h-> 0

This simplifies to

f' '(81) = lim 1 / [ ( √81+h ) + 9 ] which approaches 1/18 as h -> 0

Done.

Note that the slope of the tangent line to f(x) = √x at x = 81 is f'(81). By the definition of the derivative:
f'(81) = lim (h-->0) [f(81 + h) - f(81)]/h.
(Note that f'(x) = lim (h-->0) [f(x + h) - f(x)]/h and not lim (h-->0) [f(x + h) - f(x)]/x.)

Since f(x) = √x, we see that:
f'(81) = lim (h-->0) [f(81 + h) - f81]/h
= lim (h-->0) [√(81 + h) - √81]/h
= lim (h-->0) [√(81 + h) - 9]/h.

Rationalizing the numerator yields:
f'(81) = lim (h-->0) [√(81 + h) - 9]/h
= lim (h-->0) [(81 + h) - 81]/{h[√(81 + h) + 9]}
= lim (h-->0) h/{h[√(81 + h) + 9]}
= lim (h-->0) 1/[√(81 + h) + 9]
= 1/(9 + 9)
= 1/18.

I hope this helps!