Compute each of the following limits if they exist.
lim 3x^(2)-x-2
x->2
lim (x^(2)+x-6)/(x-2)
x->2
#1: easy, since polynomials are continuous. f(2) = 8, which is the limit
#2:
x^2+x-6 = (x+3)(x-2)
so, f(x) = x+3 everywhere except at x=2. There it is undefined, but the limit is 5 from both sides.
To compute each of these limits, we can use direct substitution. We substitute the given value of x into the expression and evaluate it.
Let's start with the first limit:
lim (3x^2 - x - 2)
x->2
To find the limit, substitute the value x = 2 into the expression:
lim (3(2)^2 - 2 - 2)
x->2
Simplifying further:
lim (3(4) - 2 - 2)
x->2
lim (12 - 2 - 2)
x->2
lim 8
x->2
The limit exists and is equal to 8.
Now, let's move on to the second limit:
lim [(x^2 + x - 6)/(x - 2)]
x->2
Again, substitute x = 2 into the expression:
lim [(2^2 + 2 - 6)/(2 - 2)]
x->2
Simplifying further:
lim [(4 + 2 - 6)/(0)]
x->2
lim [0/0]
x->2
Here, we have an indeterminate form of 0/0, which means we need to apply further calculations to find the limit. We can factorize the numerator and try canceling out the common factor with the denominator:
lim [((x - 2)(x + 3))/(x - 2)]
x->2
Now, we can cancel out the common factor:
lim (x + 3)
x->2
Substituting x = 2:
lim (2 + 3)
x->2
lim 5
x->2
The limit exists and is equal to 5.
In summary:
lim (3x^2 - x - 2)
x->2 = 8
lim [(x^2 + x - 6)/(x - 2)]
x->2 = 5