Understanding Engineering Mathematics

The table of values suggests that asx→1,f(x)→2 (the arrow represents ‘tends to’).
However, of course,

f( 1 )=

12 − 1

1 − 1

=

0

0

which isindeterminateindicating that in this case we cannot evaluate the limit simply
by substituting the value ofx=1. Instead, we proceed as follows:

lim

x→ 1

x^2 − 1

x− 1

=lim

x→ 1

(x− 1 )(x+ 1 )

(x− 1 )

=lim

x→ 1

(x+ 1 )= 2

The cancellation of thex−1 factor under the limitispermissible because in the limit we
only consider values ofxarbitrarily closebut neverequalto 1. Sox−1isneverzero
and so can be cancelled under the limit. See Figure 14.2.

y

2

01 x

y = xx^2 −− 11

The point (1,2) is

omitted from the

curve.

Figure 14.2The graph ofy=

x^2 − 1

x− 1

.

Problem 14.4

Investigate the limit off.x/=

1

x− 1

asxtends to 1.

The graph of this function is shown in Figure 14.3 and it is clear that the limit atx= 1
does not exist. If we approachx=1 from below the limit tends to−∞. If we approach
from above it tends to+∞.Wesaythereisadiscontinuityatx=1.

y

01 x

Figure 14.3The graph ofy=
1
x− 1
.