Understanding Engineering Mathematics

q

B A

P

1

Figure 14.4

From Figure 14.4 we see that ifPAis an arc of a unit circle subtending an angleθ

at the centre (173

➤

), then we have:

PB

1

=sinθ

andPA= 1 ×θ(rads).

So

sinθ

θ

=

PB

PA

and clearly asθ→0,PB→PA,so

sinθ

θ

→ 1

Thus we obtain the given result. This limit implies that forθsmall we have sinθ θ.

The same sort of reasoning can also be used to show that for very smallθ,cosθ 1.

It is also useful to note that providedk

=0,

lim

θ→ 0

sinkθ

kθ

= 1

So, for example

lim

θ→ 0

sin 6θ

6 θ

= 1

Once again, remember that in all of this discussion, and in use of the limits defined,

qmust be in radians.

(ii) lim

x→a

(

xn−an

x−a

)

=nan−^1

Whenx=athis has the form 0/0. To find the value of the limit asx→awe consider

values ofxclose toaand therefore takex=a+hso asx→a,h→0. Using the

binomial theorem (71

➤

), we have

lim

h→ 0

[

(a+h)n−an

a+h−a

]

=lim

h→ 0







an+nan−^1 h+

n(n− 1 )

2

an−^2 h^2 +···−an

h







=lim

h→ 0

[

nan−^1 +

n(n− 1 )

2

an−^2 h+···

]

=nan−^1

This result is essentially the differentiation ofxnfrom first principles (231

➤

).

(iii) limx→∞

ex

xn

=limx→∞

[

1

xn

(

1 +x+

x^2

2!

+···+

xn

n!

+···

)]

(126

➤

)

=limx→∞

(

1

xn

+

1

xn−^1

+···+

1

n!

+

x

(n+ 1 )!

+···

)

=∞