Understanding Engineering Mathematics

2. On my calculator

√

2 = 1 .41421356237. My calculator squares this to 1.9999999998,

so it is clearly not the square root of 2.

14.2 Limits

Suppose we are trying to find the velocity of a particle at a timet. As velocity is distance
moved divided by time taken we have to measure a small distance moved and divide by
the time taken for this. But this will only give us theaverage velocityover this small
interval. Actually, at the pointtthere is zero movement in zero time – is the velocity 0/0?
Remember that in Chapter 1 we said this is not defined (7
➤
). The way out of this is to
consider smaller and smaller intervals and find the average velocity over these intervals. As
the interval decreases, the average velocity over it should be nearer to the actual velocity
att. We say that we find thelimit of the velocity as the interval goes to zero. Notice
that we never actually evaluate the velocityatthe pointt– because this just gives 0/0,
as seen above – we just get nearer and nearer to it without actually reaching it. This is
formalised in the following definition.
Thelimit off.x/asxtends toais defined as the value off(x)asxapproaches
closer and closer toawithout actually reaching it and is denoted by:

lim

x→a

f(x)

There are some points to note:

(i) We do not evaluate the limit by actually substitutingx=ainf(x)

in general, although in some simple cases it is possible.

(ii) The value of the limit can depend on ‘which side it is ap-

proached’ – from ‘above’ or ‘below’, i.e. through values ofxless

thanaor through values ofxgreater thanarespectively. The two

possible values may not be the same – in which case the limit does

not exist.

(iii) The limit may not exist at all – and even if it does it may not be

equal tof(a).

The sort of limits that cause most difficulty and which are probably the most important
are those arising from so called ‘indeterminate forms’. Any expression that yields results
of the form 0/0,∞/∞,or0×∞is called anindeterminate form.Examplesare

x^2 − 1

x− 1

=

0

0

atx=1and

sinx

x

=

0

0

atx= 0

Even though the function does not exist at such points, itslimitat the point may exist.
For example as we will see below,

lim

x→ 1

(

x^2 − 1

x− 1

)

and lim

x→ 0

(

sinx

x

)

both exist.