Find Limit : Let F(x) = ( x^2 + 5; x does not equal 1)(1; x=1)
as x->c, the limit is c^2+5
However, the limit as x->1 is 6, but F(1) = 1, so F is not continuous there.
So the limit of the function as x approaches c is six for the first set and the second one f(1) equals one does not have a limit?
To find the limit of F(x) as x approaches a certain value, we need to evaluate the function as x gets closer and closer to that value. In this case, we are looking for the limit of F(x) as x approaches 1.
Since F(x) is defined differently for x = 1 and x ≠ 1, we need to consider these separately.
When x ≠ 1, F(x) is given by (x^2 + 5). To find the limit of this expression as x approaches 1, we substitute x = 1 into the expression:
lim(x→1) (x^2 + 5) = (1^2 + 5) = 6
Therefore, when x ≠ 1, the limit of F(x) is 6.
When x = 1, F(x) is given by 1. Since there are no other values of x involved, the limit is simply equal to the function value at x = 1:
lim(x→1) (1) = 1
Therefore, the limit of F(x) as x approaches 1, regardless of whether x = 1 or not, is 1.
In summary, the limit of F(x) as x approaches 1 is 1.