Use of L'Hopital's rule to evaluate Indeterminate forms:
Quote: "Evaluating indeterminate forms
The indeterminate nature of a limit's form does not imply that the limit does not exist, as many of the examples above show. In many cases, algebraic elimination, L'Hôpital's rule, or other methods can be used to manipulate the expression so that the limit can be evaluated.
For example, the expression x2/x can be simplified to x at any point other than x = 0. Thus, the limit of this expression as x approaches 0 (which depends only on points near 0, not at x = 0 itself) is the limit of x, which is 0.
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Note the phrases
"In many cases, algebraic elimination, L'Hôpital's rule, or other methods can be used to manipulate the expression"
'manipulate the expression' is the key phrase here.
"(which depends only on points near 0, not at x = 0 itself)"
'not at x = 0' is the key phrase here.
Quote: " the graph goes through ( 2, 5 ).
"
Which means that for x = 2, f(x) = 5
Let f(x) = ((x+3)*(x-2))/(x-2)
= (5*0)/0
= 0/0
Which means f(x) evaluates to 0/0 and not 5
(Note that we are not taking the limit of the function f(x) as x->2
but trying to evaluate f(x) at x=2)
The "graph would go through (2,5) for the linear equation
f(x) = x + 3 which is the graph you see.
