How do you use the limit definition to find the slope of the tangent line to the graph #3x^2-5x+2# at x=3?
Do a lot of algebra after applying the limit definition to find that the slope at
The limit definition of the derivative is:
If we evaluate this limit for
With that said, let's get started:
Evaluating this limit at
Now that we have the derivative, we just need to plug in
See the explanation section below if your teacher/textbook uses
Some presentations of calculus use, for the defintion of the slope of the line tangent to the graph of
(For example James Stewart's 8th edition Calculus p 106. On page 107, he gives the equivalent
With this definition, the slope of the tangent line to the graph of
# = lim_(xrarr3)(3x^2-5x+2-27+15-2)/(x-3)#
# = lim_(xrarr3)(3x^2-5x-12)/(x-3)#
Note that this limit has indeterminate form
# = lim_(xrarr3)(cancel((x-3))(3x+4))/cancel((x-3))#
# = lim_(xrarr3)(3x+4) = 3(3)+4 = 13#.
The limit is
#13#, so the slope of the tangent line at #x=3#is #13#.