Understanding Engineering Mathematics

The properties of limits are fairly well what we might expect. Thus, if

lim

x→a

f(x)=b lim

x→a

g(x)=c

then
L1. lim
x→a
kf (x)=kb for any fixed k

L2. xlim→a[f(x)±g(x)]=xlim→af(x)±xlim→ag(x)=b±c

L3. lim

x→a

[f(x)g(x)]=lim

x→a

f(x)lim

x→a

g(x)=bc

L4. lim

x→a

[

f(x)

g(x)

]

=

limx→af(x)

limx→ag(x)

=

b

c

ifc

= 0

L5. lim

x→a

[f(x)]

1

n=b

1

nprovided [f(x)]

1

nandb

1

nare real.

NB. We must always check that limx→af(x), limx→ag(x)exist before applying the above
results, all of which are proved rigorously in pure maths books, but are fairly ‘obvious’.

Problem 14.5

Evaluate the limits

(i) lim

x→ 1

[

x^2 − 1

x− 1

×.x^2 Y 2 xY 3 /

]

(ii) lim

x→ 1

[

x^2 − 1

.x− 1 /

√

x^2 Y 2 xY 3

]

We know from Problems 14.2 and 14.3 that

lim

x→ 1

x^2 − 1

x− 1

= 2 and lim

x→ 1

(x^2 + 2 x+ 3 )= 6

(i) Using rule L3, we have

lim

x→ 1

[

x^2 − 1

x− 1

×(x^2 + 2 x+ 3 )

]

=lim

x→ 1

x^2 − 1

x− 1

×lim

x→ 1

(x^2 + 2 x+ 3 )

= 2 × 6 = 12

which you can see we also get from:

lim

x→ 1

[(x+ 1 )×(x^2 + 2 x+ 3 )]= 12

(ii) Using rules L4 and L5, we have

lim

x→ 1

[

x^2 − 1

(x− 1 )

√

x^2 + 2 x+ 3

]

=lim

x→ 1

x^2 − 1

x− 1

÷lim

x→ 1

√

x^2 + 2 x+ 3

= 2 ÷

√

lim

x→ 1

x^2 + 2 x+ 3

= 2 ÷

√

6 =

2

√

6

=

√

6

3