There are a few variations on the limit definition of a derivative that are all equivalent, but the one most likely to be of use here is the following one:
#f'(x) = lim_{h->0} (f(x+h) - f(x)) / h#
To use that you need to evaluate (aka "plug in") both #(x+h)# and #x# into the function #f(x)#, and attempt to reduce what comes out of that, usually by cancellation, before attempting to evaluate the limit.
For instance:
#lim_{h->0} (f(x+h) - f(x)) / h #
# = lim_{h->0} ( (2(x+h)^2 - 3(x+h) + 5) - (2x^2 - 3x + 5) )/ h#
# = lim_{h->0} ( (2(x^2 + 2xh + h^2) - 3(x+h) + 5) - (2x^2 - 3x + 5) )/ h#
# = lim_{h->0} ( (2x^2 + 4xh + 2h^2 - 3x -3h + 5) - (2x^2 - 3x + 5) )/ h#
# = lim_{h->0} ( 2x^2 + 4xh + 2h^2 - 3x -3h + 5 - 2x^2 + 3x - 5 )/ h#
# = lim_{h->0} ( color(red)cancel(2x^2) + 4xh + 2h^2 color(blue)cancel(- 3x) -3h color(green)cancel(+ 5) color(red)cancel(- 2x^2) color(blue)cancel( +3x) color(green)cancel(- 5) )/ h#
# = lim_{h->0} (4xh + 2h^2 -3h) / h = lim_{h->0} (h(4x + 2h -3)) / h#
# = lim_{h->0} (4x + 2h - 3) = 4x - 3" " [A] #
In the last cancellation step [A] we are able to factor and cancel the #h# terms from the numerator and denominator because in a limit we are examining when #h# approaches 0, not when it is 0.
Once you learn basic derivatives in Calculus (or end-year Precalculus), you can verify that this is indeed the derivative of #2x^2 - 3x +5# based upon the "Power Rule" of derivatives.
