连续函数的代数–连续性和可微性12年级数学

连续函数的代数处理方程中连续函数的利用，该方程涉及您已经研究过的各种二元运算。我们还将提到一个合成规则，您可能不熟悉它，但是对于将来的应用程序来说非常重要。

由于函数在某种程度上的连续性完全取决于该函数的极限，因此合理地预期结果类似于极限情况是合理的

假设f(x)和g(x)是两个连续函数，其目的是x = a。然后我们有以下规则：

• f + g在x = x0处连续，
• f – g在x = x0处连续，
• F 。 g在x = x0处连续，并且
• f / g在x = x0(g(x)≠0)处是连续的。
• 关于连续性的复合函数定理。

如果f在g(x0)处连续且g在x0处连续，则雾在x0处连续。

两个连续函数的加法和减法

Addition of Continuous Function, 

f + g is continuous at x = x0,

Subtraction of Continuous Function, 

f – g is continuous at x = x0

证明

我们必须确定在x = a时(f(x)+ g(x))的连续性。

因此，我们需要检查要满足的三个连续性条件。由于函数f(x)和g(x)在x = a处是连续的，因此要满足连续性的所有三个条件，即

f(a) and g(a) are defined

lim_x→a f(x) = f(a) = k1 (say) and 

lim_x→ag(x) = g(a) = k2 (say)

Using them, we will get:

=> [f(a) + g(a)] is clearly defined at x = a because both f(a) and g(a) are defined.

=> Using the Summation Law of limits i.e. The limit of a sum is that the sum of the limits; 

we will get:

lim_x→a [f(x) + g(x)] = lim_x→a f(x) + lim_x→a g(x) = k1 + k2 (here)

=> f(a) + g(a) = k1 + k2 = lim_x→a [f(x) + g(x)]

因此，函数[f(x)+ g(x)]在x = a处是连续的。减法则的证明类似于加法则的证明(只需将+号替换为-号)。

两个连续函数的乘法和除法

Multiplication of Continuous Function, 

f . g is continuous at x = x0,

Division of Continuous Function, 

f/g is continuous at x = x0 (g(x) ≠ 0)

证明

Using the merchandise Law of limits i.e. The limit of a product is that the product of the limits; we will get:

lim_x→a [f(x) × g(x)] = lim_x→a f(x) × lim_x→a g(x) = k1 × k2 (here)

Using the Quotient Law of limits i.e. The limit of a quotient is that the quotient of the limits; we will get:

lim_x→a [f(x)/g(x)] = lim_x→a f(x)/lim_x→a g(x) = k1/k2 (here, provided k2 ≠ 0)

然后，证明将类似地遵循。现在看一看已解决的问题(1)，其中显示了其中1条规则的适用性。

组成规则

合成规则指出f(g(x))和g(f(x))在x = a处是连续的

问题1：讨论正弦函数的连续性。

解决方案：

To ascertain this we use the subsequent facts

lim_(x→0)⁡[sinx = 0]

We have not proved it, but is intuitively clear from the graph of sin x near 0. Now, observe that f (x) = sin x is defined for each real. Let c be a true number. Put x = c + h. If x → c we all know that h → 0. Therefore ,

lim_(x→c)⁡[f(x)] = lim_(x→c)⁡sinx

= lim_(h→0)⁡[sin⁡(c + h)]

=lim_(h→0)⁡[[sin⁡c cosh⁡] + [cosc sinh]]

= sinc + 0 = sinc = f(c)

Thus, lim_(x→c)⁡[f(x)]= f(c) and hence f may be a continuous function

问题2：证明f(x)= tan x定义的函数可能是连续函数

解决方案：

The function f(x) = tanx i.e. sinx/cosx. this is often defined for all real numbers such that cosx ≠ 0,

i.e., x ≠ 0, i.e., x ≠ (2n + 1)π/2

We have just proved that both sine and cosine functions are continuous. Thus, tanx being a quotient of two continuous functions is continuous wherever its is defined

问题3：证明函数f由

f(x)= | 1 – x + | x | |

其中x是任何实数，可以是连续函数。

解决方案：

Define g by g(x) = 1 – x + |x| and h by h(x) = |x| for all real x. Then,

(h o g)(x) = h(g(x))

= h(1 – x + |x|)

= |1 – x + |x|| = f(x)

Since we all know that h may be a continuous function. Hence, g being a sum of a polynomial function and therefore the modulus function is continuous. On the other hand, f being a composite of two continuous functions is continuous.