本文共 6184 字，大约阅读时间需要 20 分钟。

1、第一道题

对应的Java代码

public class MyMath {
       public static void main(String[] args) {
           System.out.println(f(-1));        System.out.println(f(0));        System.out.println(f(1));        System.out.println(f(3));    }    public static double f(double x){
           double y = (x * Math.sin(x - 3))/((x-1) * Math.pow((x - 3),2) );        return y;    }}

答案选择A

2、第二道题

Java代码 A选项：

package com.itzheng1.test3;public class MyMath {
       public static void main(String[] args) {
           double a = 0;        while (a <= Math.PI){
               System.out.println(A(a));            a = a+0.1;        }    }    public static double A(double x){
           double y = Math.sin(x) / x;        return y;    }}

x从0 到π值从1越来越接近0所以A是错误的

B选项

package com.itzheng1.test3;public class MyMath {
       public static void main(String[] args) {
           double a = 0;        while (a <= 10){
               System.out.println(B(a));            a = a+0.1;        }    }    public static double B(double x){
           double y = x * Math.sin(1/x);        return y;    }}

x从0到10，其结果的值逐渐趋向于1正确答案选择B

C选项

package com.itzheng1.test3;public class MyMath {
       public static void main(String[] args) {
           double a = 0;        while (a <= 10){
               System.out.println(B(a));            a = a+0.1;        }    }    public static double B(double x){
           double y = (1/x) * Math.sin(1/x);        return y;    }}

x从0到10其值越来越趋向于0

D选项

package com.itzheng1.test3;public class MyMath {
       public static void main(String[] args) {
           double a = 0;        while (a <= 10){
               System.out.println(B(a));            a = a+0.1;        }    }    public static double B(double x){
           double y = Math.sin(x) / x;        return y;    }}

的值越来越趋向于0

数学解析，正确答案选择B

3、第三道题

package com.itzheng1.test3;public class MyMath {
       public static void main(String[] args) {
           double a =10;        while (a >= 1){
               System.out.println(B(a));            a = a-0.1;        }    }    public static double B(double x){
           double y = 2 * Math.pow(Math.E,1/(x-1));        return y;    }}

当x为10趋向于1的时候其值不断的增大直到无穷

当x从-10到1的时候

package com.itzheng1.test3;public class MyMath {
       public static void main(String[] args) {
           double a =-10;        while (a < 1){
               System.out.println(B(a));            a = a+0.1;        }    }    public static double B(double x){
           double y = 2 * Math.pow(Math.E,1/(x-1));        return y;    }}

其值不断趋向于0

总结上述只有一个极限达到无穷，另外一个没有达到无穷因此答案选择D

4、第四道题

5、第五道题

Java代码：

package com.itzheng1.test3;public class MyMath2 {
       public static void main(String[] args) {
           double x = 0;        while (x < 20){
               System.out.println(f(x));            x = x + 0.1;        }    }    public static double f(double x){
           return  Math.pow(((x + 2)/(x - 1)),x);    }}

运行结果接近E三次方

6、第六道题

package com.itzheng1.test3;public class MyMath2 {
       public static void main(String[] args) {
           double a = 0;        while ( a < 100){
               System.out.println(f3(a));            a = a + 0.1;        }    }    public static double f1(double x){
           return x - 2;    }    public static  double f2(double x){
           return  (x * x) + 2 * x + 1;    }    public static  double f3(double x){
           return  (f2(f1(x)) / (x * x));    }}

运行结果接近1

7、第七道题

Java代码

package com.itzheng1.test3;public class MyMath3 {
       public static void main(String[] args) {
          double x = 10;       while (x > 0){
              System.out.println(f1(x));           x = x - 0.001;       }    }    public static double f1(double x){
             double y =  1 - Math.pow(Math.E,Math.tan(x))/Math.asin(x/2);          return  y;    }}

当x从10 到0的时候其值越来越接近-2所以a=-2

f(0)=a

8、第八道题

Java程序

package com.itzheng1.test3;public class MyMath3 {
       public static void main(String[] args) {
           double x = 10;        while (x > 0){
               System.out.println(f(x));            x = x - 0.01;        }    }    public static double f(double x){
           double y = (2 * ( Math.pow(Math.E,(-1/x)) )) / ( (Math.pow(Math.E,(2/x))  - (Math.pow(Math.E,(-1/x))) ) ) ;        return y;    }}

运行结果接近0

9、第九道题

Java代码

package com.itzheng1.test3;public class MyMath3 {
       public static void main(String[] args) {
           double x = 10;        while (x > 0){
               System.out.println(f(x));            x = x - 0.01;        }    }    public static double f(double x){
           double y = (  (  Math.sin(x) + x*x * Math.sin(1/x) ) / (2 + x * x) * Math.log(1+x) );        return  y;    }}

运行结果接近1/2

10、第十道题

11、第十一道题

x=1   x= - 1   x = 0

分别为x趋近于  1x     趋近于  -1x     趋近于   0

Java程序1

当中x趋近于1的时候

package com.itzheng1.test3;public class MyMath4 {
       public static void main(String[] args) {
           double x = 10;        while (x > 1){
               System.out.println(f(x));            x = x - 0.1;        }    }    public static double f(double x){
           double y = (x*x - x)/(x*x - 1) * Math.abs(x);        return y;    }}

其值趋近于1/2

Java程序2 当中x趋近于-1的时候

package com.itzheng1.test3;public class MyMath4 {
       public static void main(String[] args) {
           double x = -10;        while (x < -1){
               System.out.println(f(x));            x = x + 0.001;        }    }    public static double f(double x){
           double y = (x*x - x)/(x*x - 1) * Math.abs(x);        return y;    }}

其值趋近于正无穷

Java程序3 当中x趋近于0负的时候

package com.itzheng1.test3;public class MyMath4 {
       public static void main(String[] args) {
           double x = -10;        while (x < 0){
               System.out.println(f(x));            x = x + 0.001;        }    }    public static double f(double x){
           double y = (x*x - x)/(x*x - 1) * Math.abs(x);        return y;    }}

其值趋近于-1

Java程序4 当中x趋近于0正的时候

package com.itzheng1.test3;public class MyMath4 {
       public static void main(String[] args) {
           double x = 10;        while (x > 0){
               System.out.println(f(x));            x = x - 0.001;        }    }    public static double f(double x){
           double y = (x*x - x)/(x*x - 1) * Math.abs(x);        return y;    }}

其值取决于1

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