as p approaches 1

does not exist

the denominator is 0 when p is substituted by 1, so we cannot use the quotient limit law

if you attempted to multiply by the conjugate (of either the numerator or denominator), you would not get anywhere useful.

so let us check the left- and right- hand limits. the two-sided limit exists if and only if the one-sided limits (from the left and from the right) exist and are equal to each other.

consider the limit

if p approaches 1 but is less than one , then the numerator approaches while the denominator approaches a small positive number (since . so we have that the quotient becomes a large positive number and so

if p approaches 1 but is greater than 1 , then the numerator approaches while the denominator approaches a small negative number (since ). we have that the quotient becomes a large negative number and so

the left- and right-hand limits are different, so the limit does not exist

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)