Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

SERIES AND LIMITS


Evaluate the limits

lim
x→ 1

(x^2 +2x^3 ), lim
x→ 0

(xcosx), lim
x→π/ 2

sinx
x

.


Using (a) above,


lim
x→ 1

(x^2 +2x^3 ) = lim
x→ 1

x^2 + lim
x→ 1

2 x^3 =3.

Using (b),


lim
x→ 0

(xcosx) = lim
x→ 0

xlim
x→ 0

cosx=0×1=0.

Using (c),


lim
x→π/ 2

sinx
x

=


limx→π/ 2 sinx
limx→π/ 2 x

=


1


π/ 2

=


2


π

.


(iv) Limits of functions ofxthat contain exponents that themselves depend on

xcan often be found by taking logarithms.


Evaluate the limit

lim
x→∞

(


1 −


a^2
x^2

)x 2
.

Let us define


y=

(


1 −


a^2
x^2

)x 2

and consider the logarithm of the required limit, i.e.


lim
x→∞

lny= lim
x→∞

[


x^2 ln

(


1 −


a^2
x^2

)]


.


Using the Maclaurin series for ln(1 +x) given in subsection 4.6.3, we can expand the
logarithm as a series and obtain


lim
x→∞

lny= lim
x→∞

[


x^2

(



a^2
x^2


a^4
2 x^4

+···


)]


=−a^2.

Therefore, since limx→∞lny=−a^2 it follows that limx→∞y=exp(−a^2 ).


(v) L’Hopital’s rule may be used; it is an extension of (iii)(c) above. In casesˆ

where both numerator and denominator are zero or both are infinite, further


consideration of the limit must follow. Let us first consider limx→af(x)/g(x),


wheref(a)=g(a) = 0. Expanding the numerator and denominator as Taylor


series we obtain


f(x)
g(x)

=

f(a)+(x−a)f′(a)+[(x−a)^2 /2!]f′′(a)+···
g(a)+(x−a)g′(a)+[(x−a)^2 /2!]g′′(a)+···

.

However,f(a)=g(a)=0so


f(x)
g(x)

=

f′(a)+[(x−a)/2!]f′′(a)+···
g′(a)+[(x−a)/2!]g′′(a)+···

.
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