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The Real And Complex Number Systems

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Proof: Trivially, f x 1/x 2 1<br />

cos1/x, ifx 0. Let x n <br />

2n 1 2<br />

consider<br />

fx n f0<br />

sin1/x n<br />

x n 0 x n<br />

2n 1 as n .<br />

2<br />

Hence, we know that f 0 does not exist.<br />

(b) g x sin1/x 1/x cos1/x, ifx 0; g 0 does not exist.<br />

, and thus<br />

Proof: Trivially, g 1<br />

x sin1/x 1/x cos1/x, ifx 0. Let x n ,and<br />

2n 1 2<br />

y n 1 , we know that 2n<br />

gx n g0<br />

sin 1<br />

x n 0 xn<br />

1 for all n<br />

and<br />

gy n g0<br />

sin 1<br />

y n 0 yn<br />

0 for all n.<br />

Hence, we know that g 0 does not exist.<br />

(c) h x 2x sin1/x cos1/x, ifx 0; h 0 0; lim x0 h x does not exist.<br />

Proof: Trivially, h x 2x sin1/x cos1/x, ifx 0. Consider<br />

hx h0<br />

|x sin1/x| |x| 0asx 0,<br />

x 0<br />

so we know that h 1<br />

0 0. In addition, let x n ,andy<br />

2n 1 n 1 ,wehave<br />

2n<br />

2<br />

h x n 2<br />

2n 1 and h y n 1 for all n.<br />

2<br />

Hence, we know that lim x0 h x does not exist.<br />

5.6 Derive Leibnitz’s formula for the nth derivative of the product h of two functions<br />

f and g :<br />

n<br />

h n kn f k g nk x, where kn n!<br />

k!n k! .<br />

k0<br />

Proof: We prove it by mathematical Induction. As n 1, it is clear since<br />

h f g g f. Suppose that n k holds, i.e., h k k<br />

j0<br />

jk f j g kj x. Consider<br />

n k 1, we have

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