Understanding Engineering Mathematics

and again you can see we get this directly from

lim

x→ 1

[

x+ 1

√

x^2 + 2 x+ 3

]

=

2

√

6

Note that it wouldnotbe correct to write

lim

x→ 1

[

x^2 − 1

(x− 1 )

√

x^2 + 2 x+ 3

]

=lim

x→ 1

(x^2 − 1 )÷lim

x→ 1

((x− 1 )

√

x^2 + 2 x+ 3 )

since the last limit, which we are trying to divide by, is zero.

Exercises on 14.2

1. Investigate the (very useful) limit of

1

x

asxtends to infinity, i.e. gets infinitely large.

Evaluate

(i) lim

x→∞

1

x− 1

(ii) lim

x→∞

2 x

x− 4

2. Consider the values of the following functions asxgets closer and closer (but not
   equal) tox=2:

(i) x− 2 (ii) x+ 2 (iii) x^2 − 4

(iv)

1

x^2 − 4

(v)

x− 2

x^2 − 4

In each case evaluate the limit asx→2 using the techniques of this section.

3. Evaluate the limits

(i) lim

x→ 0

x^2 (ii) lim

x→ 0

x

x^2 −x

(iii) lim

x→ 1

x− 1

x^2 − 3 x+ 2

Answers

1. lim
   x→∞

1

x

= 0 (i) 0 (ii) 2

2. (i) 0 (ii) 4 (iii) 0 (iv) doesn’t exist (v)^14


3. (i) 0 (ii) − 1 (iii) − 1

14.3 Some important limits

There are further examples of limits that are very important in calculus and approximations.
If you wish, simply remember the results and how to use them, but working through
the proofs will give you good practice in basic mathematics as well as enhancing your
understanding of the results.

(i) lim

θ→ 0

sinθ

θ

= 1

Whenθ=0 this yields the indeterminate form 0/0.