Введение в математический анализ в вопросах и задачах. Анчиков А.М - 65 стр.

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b) |x
0
| > 0 |h| = |x − x
0
| < |x
0
|.
arctg (x
0
−h) −arctg x
0
= t, tg t =
h
1+x
2
0
+hx
0
. | t | <
| tg t | | t | <
π
2
, | arctg (x
0
− h) − arctg x
0
| < | t | <
| tg t | =
¯
¯
¯
h
1+x
2
0
+hx
0
¯
¯
¯ <
| h |
1+x
2
0
+hx
0
< ε, | h | = | x − x
0
| <
(1+x
2
0
)ε
1+| x
0
|ε
= δ
ε
(x
0
). arctg x x
0
= 0
| arctg x − arctg 0 | ≤ | arctg x | < | x |.
f(x) =
= sin
1
x
, x 6= 0 f(0)
{x
n
}
{x
0
n
} x
n
=
1
πn
, x
0
n
=
1
π(1+4n)
, (n = 1, 2, ...).
n → +∞ x
n
→ 0, x
0
n
→ 0, f(x
n
) → 0, f(x
0
n
) → 1,
lim
x→0
sin
1
x
f(x)
x = 0. x 6= 0,
f(x) =
³
1
x
−
1
x+1
´
.
³
1
x−1
−
1
x
´
x = −1, 0, +1.
lim
x→−1±0
f(x) = lim
x→−1±0
x−1
x+1
= ∓ ∞, lim
x→0
f(x) = lim
x→0
x−1
x+1
=
= −1, lim
x→1
f(x) = lim
x→1
x−1
x+1
= 0, x = 0 x = + 1
x = − 1
f(g(x)) g(f(x)) f(x) = sgnx g(x) = x ·(1 −x
2
).
f(g(x)) = sgn[x · (1 − x
2
)] =
      b) Ïóñòü |x0 | > 0 è |h| = |x − x0 | < |x0 | . Åñëè
arctg (x0 − h) − arctg x0 = t, òî tg t = 1+x2h+hx0 . Òàê êàê | t | <
                                                   0
                         π
| tg t | ïðè
           ¯
               | t | < ¯2
                           , òî  |  arctg (x 0 − h)  − arctg x0 | < | t | <
           ¯     h     ¯        |h|
| tg t | = ¯ 1+x2 +hx0 ¯ < 1+x2 +hx0 < ε, åñëè | h | = | x − x0 | <
                  0                0
(1+x20 )ε
       = δε (x0 ). Íåïðåðûâíîñòü arctg x ïðè x0 = 0 ñëåäóåò
1+| x0 |ε
èç íåðàâåíñòâà | arctg x − arctg 0 | ≤ | arctg x | < | x | .
    Ïðèìåð 50. Èññëåäîâàòü íà íåïðåðûâíîñòü ôóíêöèþ f (x) =
= sin x1 , åñëè x 6= 0 è f (0) ïðîèçâîëüíî.
    Ðåøåíèå. Ðàññìîòðèì äâå ïîñëåäîâàòåëüíîñòè          {xn } è
                       1                  1
{x0n }ãäå xn =        πn
                         ,  =x0n       (n = 1, 2, ...). Òàê êàê
                                       π(1+4n)
                                               ,
ïðè n → +∞ xn → 0, xn → 0, à f (xn ) → 0, f (x0n ) → 1, òî
                          0

ïðåäåë lim sin x1 íå ñóùåñòâóåò. Ýòî çíà÷èò, ÷òî f (x) òåðïèò
        x→0
ðàçðûâ âòîðîãî ðîäà ïðè x = 0. Åñëè x 6= 0, òî íåïðåðûâíîñòü
î÷åâèäíà.
    Ïðèìåð 51. Îïðåäåëèòü òî÷êè
                            ³   ðàçðûâà
                                    ´   è èññëåäîâàòü õà-
                                            1        1    .³                 ´
ðàêòåð ýòèõ òî÷åê, åñëè f (x) =                 −               1        1
                                            x       x+1
                                                               x−1
                                                                     −   x

    Ðåøåíèå. Ôóíêöèÿ íå îïðåäåëåíà ïðè x = −1, 0, +1. Òàê
êàê         lim f (x) =  lim x−1 = ∓            ∞, lim f (x) = lim               x−1
                                                                                       =
         x→−1±0        x→−1±0 x+1                    x→0                     x→0 x+1
                             x−1
= −1, lim f (x)       = lim x+1òî x = 0 è x = + 1 
                                  = 0,
         x→1             x→1
óñòðàíèìûå òî÷êè ðàçðûâà, à x = − 1  òî÷êà áåñêîíå÷íîãî
ðàçðûâà.
    Ïðèìåð 52. Èññëåäîâàòü íà íåïðåðûâíîñòü ñëîæíûå ôóíê-
öèè f (g(x)) è g(f (x)) åñëè f (x) = sgnx è g(x) = x · (1 − x2 ).
    Ðåøåíèå. f (g(x)) = sgn[x · (1 − x2 )] =


                                       65

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