The limit of the function as x approaches 3 for the expression x² is proven to equal 9. The proof utilizes the ε-δ definition of limits, stating that for every ε>0, there exists a δ>0 such that if 0<|x-3|<δ, then |x²-9|<ε. The solution involves manipulating the expression |x²-9| into |x-3||x+3| and applying the triangle inequality to establish bounds on |x+3|, leading to the conclusion that δ can be chosen as the minimum of 1 and ε/7.
PREREQUISITESStudents studying calculus, particularly those focusing on limits and proofs, as well as educators looking for examples to illustrate the ε-δ definition of limits.
WK95 said:Homework Statement
Prove that
##\lim_{x \rightarrow 3} x^{2} = 9##
Homework Equations
For every ε>0, there is a δ<0 so that if 0<|x-a|<δ then |f(x)-L|<ε
The Attempt at a Solution
##If~0<|x - 3|<δ~then~|x^2 - 9|<ε##
##|x^2 - 9|##
##|x - 3||x + 3|##
At this point, I would argue that if, say, |x- 3|< 1, then 2< x< 4 so that 5< x+ 3< 7.WK95 said:Homework Statement
Prove that
##\lim_{x \rightarrow 3} x^{2} = 9##Homework Equations
For every ε>0, there is a δ<0 so that if 0<|x-a|<δ then |f(x)-L|<ε
The Attempt at a Solution
##If~0<|x - 3|<δ~then~|x^2 - 9|<ε##
##|x^2 - 9|##
##|x - 3||x + 3|##
##= |x - 3||x - 3 + 3 + 3|##
##= |x - 3|*|(x - 3) + 6|##
##≤ |x - 3|*(|x - 3| + |6|), by~triangle~inequality##
##= |x - 3|2 + 6|x - 3|##
##<##