Mathematical Methods for Physics and Engineering - Matematica.NET

SERIES AND LIMITS◮Evaluate the limitslimx→1 (x2 +2x 3 ),lim(x cos x),x→0sin xlimx→π/2 x .Using (a) above,Using (b),Using (c),limx→1 (x2 +2x 3 ) = lim x 2 + lim 2x 3 =3.x→1 x→1lim(x cos x) = lim x lim cos x =0× 1=0.x→0 x→0 x→0sin xlimx→π/2 x= lim x→π/2 sin xlim x→π/2 x = 1π/2 = 2 π . ◭(iv) Limits of functions of x that contain exponents that themselves depend onx can often be found by taking logarithms.◮Evaluate the limit) x 2lim(1 − a2.x→∞ x 2Let us definelim ln y = limx→∞ x→∞) x 2y =(1 − a2x 2and consider the logarithm of the required limit, i.e.[x 2 ln(1 − a2x 2 )].Using the Maclaurin series for ln(1 + x) given in subsection 4.6.3, we can expand thelogarithm as a series and obtainlim ln y = lim[x(− ···)]2 a2x→∞ x→∞ x − a42 2x + = −a 2 .4Therefore, since lim x→∞ ln y = −a 2 it follows that lim x→∞ y =exp(−a 2 ). ◭(v) L’Hôpital’s rule may be used; it is an extension of (iii)(c) above. In caseswhere both numerator and denominator are zero or both are infinite, furtherconsideration of the limit must follow. Let us first consider lim x→a f(x)/g(x),where f(a) =g(a) = 0. Expanding the numerator and denominator as Taylorseries we obtainf(x)g(x) = f(a)+(x − a)f′ (a)+[(x − a) 2 /2!]f ′′ g(a)+(x − a)g ′ (a)+[(x − a) 2 /2!]g ′′ (a)+···.However, f(a) =g(a) =0sof(x)g(x) = f′ (a)+[(x − a)/2!]f ′′ g ′ (a)+[(x − a)/2!]g ′′ (a)+···.142