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|##
##<##
The limit of a function is the value that the function approaches as the input approaches a certain value or point.
A limit exists if the function approaches a single, finite value as the input approaches a certain value or point. In other words, the left and right limits are equal.
A limit is evaluated by plugging in values that approach the given point from both the left and right sides of the function and observing the resulting outputs. If these outputs approach the same value, then the limit exists and is equal to that value.
To prove this limit, we need to show that for any positive number ε, we can find a positive number δ such that for all x, if 0 < |x - 3| < δ, then |x^2 - 9| < ε. This can be done using the definition of a limit and algebraic manipulation to find a suitable δ value.
Yes, a limit can exist even if the function is not defined at the given point. This is because the limit only considers the behavior of the function as the input approaches the given point, not the actual value of the function at that point.