Continuity of operation and primary functions continuous function
Continuity of the four arithmetic continuous function
And the continuous function limit four operations can be obtained at a certain point the following theorems:
Theorem 1
Provided the function \ (f (x) \) and \ (g (x) \) at the point \ (x_0 \) at a continuous, their sum (difference) \ (F \ PM G \) , the product \ (F \ G CDOT \) , commercially \ (\ frac {f} { g} (g \ not = 0) \) are at a point \ (x_0 \) in continuity.
Continuous inverse function of the complex function
Theorem 2
If the function \ (y = f (x) \) in the interval (I_x \) \ monotonically increasing (monotonic decrease) and continuous, then its inverse function on \ (x = f ^ {- 1} (y) \) also in the section for \ (I_y = \ {y | y = f (x), x \ in I_x \} \) single tone incrementing u (monotonically decreasing) and continuous.
Theorem 3
Provided function \ (y = f [g ( x)] \) by a function (u = g (x) \ ) \ and a function \ (y = f (x) \) compound formed, \ (\ mathring the U-{} (x_0) \ in D_ {F \ CIRC G} \) . If \ (\ lim_ {X \ to x_0} G (X) = U_0 \) , and the function \ (y = f (u) \) in \ ( u = u_0 \) continuously is $$ \ lim_ {x \ to x_0 } f [g (u)] = \ lim_ {u \ to u_0} f (u) = f (u_0). $$
Proof: In the composite function limits the algorithm , so that \ (A = F (U_0) \) (time \ (f (u) \) at the point \ (U_0 \) continuous), and Cancel "present \ (\ delta_0 > 0 \) when \ (x \ in \ mathring { U} (x_0, \ delta_0) \) when there is \ (G (X) \ = U_0 Not \) "the condition, have over the above theorem, where \ (g (x) \ not = u_0 \) reason for this condition can be canceled are: \ (\ FORALL \ varepsilon> 0 \) , so \ (g (x) = u_0 \) those points established \ (x \ ) , apparently the \ (| f [g (x )] - f (u_0) | <\ varepsilon \) established therefore additional. \ (G (the X-) \ not = U_0 \) this condition is also not necessary.
Since there Theorem 3 $$ \ lim_ {x \ to x_0 } f (x) = u_0, \ lim_ {u \ to u_0} f (u) = f (u_0), $$ so $$ \ lim_ {x \ to x_0} f [g (u )] = \ lim_ {u \ to u_0} f (u) = f (u_0) $$ can be expressed as $$ \ lim_ {x \ to x_0 } f [g (x)] = f [\ lim_ {x \ to x_0} g (x)]. $$ represented by the formula, under the conditions of Theorem 3, if for substitution \ (U = F (X) \) , then find \ (\ lim_ {x \ to x_0} f [g (x)] \) turned into request \ (\ lim_ {U \ to U_0} F (U) \) , where \ (u_0 = \ lim_ {x \ to x_0} G (X) \) . under the conditions of Theorem 3, seeking composite function \ (f [g (x) ] \) at the limit, function symbols \ (F \) and limit the symbol \ (\ lim_ {x \ to x_0} \) can be exchanged order.
Theorem in the 3 \ (x \ to x_0 \) into \ (X \ to \ infty \) , gives similar theorem.
Theorem 4
** set function \ (y = f [g ( x)] \) is a function (u = g (x) \ ) \ and a function \ (y = f (u) \) compound formed, \ (the U-( x_0) \ Subset D_ {F \ CIRC G} \) . If the function \ (u = g (x) \) in \ (x = x_0 \) continuously, and \ (G (x_0) = U_0 \) , and the function \ (y = f (x) \) in \ (u = u_0 \) continuously, in line with the function \ (y = f [g ( x)] \) in \ (x = x_0 \) is also continuous.
Proof: Theorem 3 in order as long as the \ (U_0 = G (x_0) \) , which represents \ (g (x) \) at the point \ (x_0 \) continuously, so $$ \ lim_ {x \ to x_0 } f [g (u)] = \ lim_ {u \ to u_0} f (u) = f (u_0) $$ obtain $$ \ lim_ {x \ to x_0 } f [g (x)] = f (u_0) = f [g (x_0)] , $$ which proves the compliance function (f [g (x)] \) \ point (x_0 \) \ continuity.