Funksiyani hosila yordamida tckshirish va grafigini

8.9.1-misol. Ushbu f(x)=arctgx funksiyani 0(xⁿ) hadgacha Makloren formulasiga yoying:
Yechilishi. Ma’lumki,

f ⁿ (x)
(n  1)!
 sin ^n(arctg x  ^ )^,

(1  x² )^{n/2 }_
2 _

(n)

_0, agarda n juft bo 'lsa ,

f (0)  _
_


(1)
т1 2

(т  1)! , agarda n toq bo 'lsa.

Demak, Peano ko‘rinishidagi qoldiq hadli Makloren formulasiga asosan, (5)

3
arctg x  x  ^x
3

• 
  _x5
  

5

• 
  _x7
  

7
^n1 _xn
 ...  (1) ²  0(xⁿ )
n
(26)

10. Teylor formulasidan foydalanib limitlarni hisoblash

Ushbu
_lim f (x)
^x0 g(x)
limitni topish talab qilingan bo‘lsin, bunda

f(0)=g(0)=0. f(x) va g(x) lar Makloren formulasiga yoyiladigan funksiyalar bo‘lsin. Bu funksiyalarning Makloren formulasidagi yopilmalarining birinchi noldan farqli hadi bilan chegaralanamiz:

Agar m=n bo‘lsa,
f (x)  axⁿ  0(xⁿ ) ,
g(x)  bx^m  0(x^m ) ,
a  0,
b  0.

lim
f (x)


 lim
axⁿ  0 (xⁿ ) _ a
(1)

bo‘ladi.
Agar

bo‘ladi.
n  m

bo‘lsa,
^x0 g(x)

_lim f (x)
^x0 g(x)
^x0 bx^m  0 (x^m ) b

lim _{m m}

0
_ axⁿ  0 (xⁿ ) _
^x0 bx  0 (x )

2. 
   

Agar
n  m bo‘lsa,
lim


f (x)
 lim


axⁿ  0 (xⁿ ) _{ }

bo‘ladi.
^x0 g(x)

^x0 bx^m  0 (x^m )

3. 
   

8.10.1-misol. Ushbu


_lim tg x  sin x

limitni toping.

Yechilishi.

tg x 

 ^x3 

x
3
0 (x⁴ )
x0 _x³

,

3
sin x  x  ^x
6
 0(x⁴ )

yoyilmalardan foydalanib, berilgan limitni topamiz:

tg x 
¹ x³  0 (x ⁴ ) _{ }

lim ^{sin x}  lim ²  lim_¹  ^0(x
) _{ } 1 _.

x0 _x³
x0 _x³
x0 _{ 2}
x³ _ 2