CK-12-Pre-Calculus Concepts

http://www.ck12.org Chapter 14. Concepts of Calculus

b. limx→ 2 +(x−^12 )

c. limx→ 1 +

(x (^2) + 2 x− 3
x− 1

)

Solution:Most of the time one sided limits are the same as the corresponding two sided limit. The exceptions are
when there are jump discontinuities, which normally only happen with piecewise functions, and infinite discontinu-
ities, which normally only happen with rational functions.
a. limx→ 3 −( 4 x− 3 ) = 4 · 3 − 3 = 12 − 3 = 9

b. limx→ 2 +(x−^12 )=DNEor∞

The reason why∞is preferable in this case is because the two sides of the limit disagree. One side goes to negative
infinity and the other side goes to positive infinity (see the graph below). If you just indicate DNE then you are
losing some perfectly good information about the nature of the function.

c. limx→ 1 +

(x (^2) + 2 x− 3
x− 1

)

=xlim→ 1 +

((x− 1 )(x+ 3 )

(x− 1 )

)

=xlim→ 1 +(x+ 3 ) = 1 + 3 = 4

Concept Problem Revisited
In order to confirm or deny that the function is continuous, graphical tools are not accurate enough. Sometimes jump
discontinuities can be off by such a small amount that the pixels on the display of your calculator will not display a
difference. Your calculator will certainly not display removable discontinuities.

f(x) =







−x− 2 x< 1

− 3 x= 1

x^2 − 4 1 <x

You should note that on the graph, everything to the left of 1 is continuous because it is just a line. Next you should
note that everything to the right of 1 is also continuous for the same reason. The only point to check is atx=1. To
check continuity, explicitly use the definition and evaluate all three parts to see if they are equal.

• limx→a−f(x) =− 1 − 2 =− 3


• f( 1 ) =− 3


• limx→ 1 +f(x) = 12 − 4 =− 3

Therefore, limx→ 1 −f(x) =f( 1 ) =xlim→ 1 +f(x)and the function is continuous atx=1 and everywhere else.