082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

10.3 Limits and continuity 359
f (t)
5
y = 1 – x
y
4
3
y = x +1
2
–3 2 t
Figure10.5
Thecurvef(t) =t 2 +2t −3.
–1
Figure10.6
Asx → 0,y → 1,even though
y(0) =3.
1
0
3 x
Example10.1 Ift → 2,whatvalue does
approach?
f(t)=t 2 +2t−3
Solution Figure 10.5 shows a graph of f (t). Clearly, whethert = 2 is approached from the l.h.s.
orther.h.s.thefunctiontendsto5.Thatis,ift → 2,then f (t) → 5.Wenotethatthisis
thevalueof f (2).Informallywearesayingthatast getsnearerandnearertothevalue2,
so f (t)gets nearerand nearer to5.This isusuallywritten as
lim(t 2 +2t−3)=5
t→2
where ‘lim’ is an abbreviation of limit. In this example, the limit of f (t) ast → 2 is
simply f (2), butthisisnottrueforall functions.
Example10.2 Figure 10.6 illustratesy(x) defined by
⎧
⎨1−x x<0
y(x) = 3 x=0
⎩
x+1 x>0
Evaluate:
(a) lim x→3
y (b) lim x→−1
y (c) lim x→0
y
Solution We notethat thisfunctionispiecewise continuous.Ithas a discontinuityatx = 0.
(a) Weseekthelimitofyasxapproaches3.Asxapproaches3,wewillbeonthatpart
of the function defined byx > 0, that isy(x) =x +1. Asx → 3, theny → 3 +1,
that isy → 4.So
lim y = 4
x→3
(b) Whenxapproaches −1, we will be on that part of the function defined byx < 0,
thatisy(x) = 1 −x.Soasx → −1,theny → 1 − (−1),thatisy → 2.Hence
lim y = 2
x→−1
(c) Asxapproaches0whatvaluedoesyapproach?Notethatwearenotevaluatingy(0)
which actually has a value of 3. We simply ask the question ‘What value isynear