QUESTION 18
a) 𝑙𝑖𝑚 𝑠𝑞𝑟𝑡(𝑥 − 3) 𝑎𝑠 (𝑥 −> 3)
This limit does not exist, mainly because as x approaches 3, the function (sqrt(x-3) ) becomes 0. If we
then take the square root of 0 we get 0. However, the function is undefined for x values less than 3
and as such the limit cannot be evaluated.
𝑥 2 −4
b) 𝑙𝑖𝑚 𝑥 2 −4𝑥 +4 𝑎𝑠 (𝑥−> 2)
The given limit does not exist simply because as x approaches 2, the denominator (𝑥^2 − 4𝑥 +
4) becomes 0 resulting in the creation of an indeterminate form 0/0. As such, as it stands the limit of
this function does not exist till further factoring is done
c) 𝑓(𝑥) = (−5, 𝑖𝑓 𝑥 < 1) (2, 𝑖𝑓 𝑥 >= 1) 𝑙𝑖𝑚 𝑓(𝑥) 𝑎𝑠 𝑥−> 1
The given limit only exists for (2, if x>=1) as x-> 1. This is because, when f(x) = (-5, if x<1) the function is defined as -5 for x
values that are less than 1, there is a jump discontinuity when x = 1. The left-hand limit as x goes to 1 is -5, but the righthand limit as x reaches 1 is 2. Since the two limits are not equal the limit of f(x) as x approaches 1 does not exist.
d) 𝑙𝑖𝑚(1)/(𝑠𝑞𝑟𝑡(𝑥 − 2)) 𝑎𝑠 𝑥 −> 2
The given limit does not exist because as x approaches 2 the expression inside the square root (x-2) becomes 0 and the
square root of 0 makes the denominator 0 resulting in an undefined value. Therefore, the limit is undefined.
e) 𝑙𝑖𝑚
(𝑎𝑏𝑠(𝑥))
𝑥
𝑎𝑠 𝑥−> 0
the given limit does not exist because as x approaches 0, the expression (abs(x)) approaches 0 while x itself approaches 0
since the limit of the numerator and denominator are not the same, the limit is undefined
f) 𝑓(𝑥) = (5𝑥 2 , 𝑖𝑓 𝑥 < −1), (2𝑥 + 1, 𝑖𝑓 𝑥 >= −1) 𝑙𝑖𝑚 𝑓(𝑥) 𝑎𝑠 𝑥−> −1
The given limit does exist but only for the function (2x+1) because as x approaches -1 for the function 5𝑥 2 there is a jump
discontinuity at x = -1, and the left-hand limit is different from the right hand limit.