Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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)+···

.