Using the limit definition, how do you differentiate #f(x) = x^2 - 1598#?

1 Answer
Jan 6, 2016

Expand and simplify. Then evaluate the limit.

Explanation:

#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

For this function, we get

#f'(x) = lim_(hrarr0)([(x+h)^2-1598] - [x^2-1598])/h#

If we try to evaluate the limit by substitution, we get the indeterminate form #0/0#, so we need a different approach.
It may not be clear that expanding the numerator will help, but we need to try something, so let's do it and see what happens.

#f'(x) = lim_(hrarr0)([x^2+2xh+h^2-1598] - [x^2-1598])/h#

# = lim_(hrarr0)(color(red)(x^2)+2xh+h^2color(blue)(-1598) color(red)(-x^2)color(blue)(+1598))/h#

# = lim_(hrarr0)(2xh+h^2)/h#

If we try substitution, we still get indeterminate form #0/0#.

Note however, that for all #h# other that #0#, we have

#(2xh+h^2)/h = (cancel(h)(2x+h))/cancel(h) = 2x+h#

Therefore,

#f'(x) = lim_(hrarr0)(2xh+h^2)/h = lim_(hrarr0)(2x+h)=2x#

In many classes it is permissible to write this without comment as follows:

#f'(x) = lim_(hrarr0)([(x+h)^2-1598] - [x^2-1598])/h#

#= lim_(hrarr0)([x^2+2xh+h^2-1598] - [x^2-1598])/h#

# = lim_(hrarr0)(x^2+2xh+h^2-1598 -x^2+1598)/h#

# = lim_(hrarr0)(2xh+h^2)/h#

# = lim_(hrarr0)(2x+h)#

# = 2x#