Конспект лекций по математическому анализу. Шерстнев А.Н. - 63 стр.

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G0(y) 6= 0. tEPERX
             lim F (y) = xlim
            y!0            !1 f (x) = 0; ylim !0
                                                 G(y) = xlim
                                                          !1 g (x) = 0;
                              f 0 ( 1 )(, 1 )
           lim  f (x) = lim y y2 = lim F 0(y) = lim F (y)
                  0
          x!1 g 0(x)     y!0 0 1                    0
                              g ( y )(, y12 ) y!0 G (y) y!0 G(y)
                         = xlim     f (x)
                             !1 g (x)
(W PREDPOSLEDNEM RAWENSTWE MY WOSPOLXZOWALISX UVE RAZOBRANNYM SLU-
^AEM (0=0) DLQ SOBSTWENNOJ TO^KI a = 0).
    50: (1=1); U = (a; ); x ! a+. pUSTX xlim     f 00(x) = . dLQ x,
                                              !a+ g (x)
DOSTATO^NO BLIZKIH K a, IMEEM (S U^ETOM TEOREMY kO[I DLQ OTREZKA
[x; ]  (a; ))
                f (x) = f (x)[g() , g(x)]  f () , f (x) = h(; x) f 0(c)
                g(x) g(x)[f () , f (x)] g() , g(x)                          g0(c)
GDE x < c <  I h(; x) = [1 , ff((x)) ],1(1 , gg((x)) ). pARAMETROM  W PRAWOJ
^ASTI RAWENSTWA MY MOVEM                 RASPORQVATXSQ. pUSTX " > 0 PROIZWOLXNO
                                   0 (y ) f 0 (x)
                                 f
(" < 1) I  TAKOWO, ^TO j g0(y) , g0(x) j  "=3 DLQ WSEH x; y 2 (a; ). w SILU
USLOWIQ (1=1) SU]ESTWUET N > 0 TAKOE, ^TO
                        0
             jh(; x) fg0((xx)) , j < "=3; jh(; x) , 1j < "=3 (jxj > N ):
tOGDA DLQ jxj > N
                                                 0     0              0            0
       j fg((xx)) , j = j(h(; x) , 1)( fg0((cc)) , fg0((xx)) ) + ( fg0((cc)) , fg0((xx)) )
                                         0
                          + h(; x) fg0((xx)) , j < ";

TO ESTX xlim f 00(x) = : >
         !a+ g (x)


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