What is the domain of f(x)= (x-3)/(x^2-9)?
lim f(x)
x->3
domain of f(x) :
any real number, x ≠ 3
lim (x-3)/(x^2 - 9) , x --->3
= lim (x-3)((x-3)(x+3))
= lim 1/(x+3) , as x -->3
= 1/6
x = -3 is also excluded from the domain, eh? since x^2-9 = 0
To find the domain of a function, you must identify any values of x that would make the function undefined.
In this case, the function f(x) = (x-3)/(x^2-9) will be undefined if the denominator, x^2-9, is equal to zero.
To find the values of x that make the denominator zero, we can set x^2-9 = 0 and solve for x.
x^2 - 9 = 0
(x-3)(x+3) = 0
From the above equation, x can either be 3 or -3.
Therefore, the domain of f(x) is all real numbers except x = 3 and x = -3. In interval notation, the domain can be expressed as (-∞, -3) U (-3, 3) U (3, ∞).
Now, let's find the limit of f(x) as x approaches 3.
lim f(x)
x->3
In this case, it is possible to directly substitute x = 3 into the function since 3 is not in the domain exclusion set.
lim (x-3)/(x^2-9)
x->3
Plugging in x = 3, we get:
lim (3-3)/(3^2-9)
x->3
Simplifying further:
lim 0/(9-9)
x->3
Since the numerator is zero, and the denominator is also zero, we have an indeterminate form of 0/0.
To evaluate this limit further, we can apply algebraic manipulation or use techniques such as factoring, rationalizing, or L'Hopital's rule depending on the complexity of the function.
However, in this case, we can simplify the expression:
lim 0/(9-9)
x->3
lim 0/0
x->3
Since 0 divided by anything is always 0, we can conclude that the limit of f(x) as x approaches 3 is 0.