Apply the precise definition of limit to show that limit as x approaches 4 of x^(2) -4x + 5 =1

Answer 1

x^2 -4x+5 approaches 5 as x approaches 4

4^2-4(4) +5 = 16-16+5= 0+5=5

graph y=x^2-4x+5

it's a parabola

y= x^2-4x+4 +5-4

= (x-2)^2 +1 with vertex = minimum = (2,1)

when x=4, y = (4-2)^2+1 = 2^2+1 = 4+1=5

for any epsilon e >0, there is a delta d>0

whenever |f(x)-L|< e, there exists d such that 0<|x-a|< d

limit of y is the tangent = slope of the graph of the equation

as x approaches 4

tangent line's slope approaches 5 as x approaches 4