用Octave判断间断点的分类（1）
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（1）可去间断点
设\( f(x) = \cases{ \frac{ {\rm ln}(1+x)}{x} , & x > 0 \\ 0 , & x = 0 \\ \frac{ \sqrt{1+x} - \sqrt{1-x} }{x} , & -1 ≤ x < 0 } \)，讨论\( f(x) \)在\( x=0 \)处的连续性。
程序代码如下
function [text_result, numeric_result] = func15(x_value)
    pkg load symbolic;
    x = sym('x');
    equation_1 = log(1 + x) / x;
    equation_2 = 0;
    equation_3 = (sqrt(1 + x) - sqrt(1 - x)) / x;
    if (x_value > 0)
        question = equation_1;
    elseif (x_value == 0)
        question = equation_2;
    elseif (x_value >= -1)
        question = equation_3;
    else
        error('Invalid input value.')
    endif
    lim = limit(question, x, x_value);
    text_result = ["\n", disp(lim)];
    numeric_result = eval(lim);
endfunction
令\(x \gt 0 ( x = 0.01)\)，计算结果如下
>> [text_result, numeric_result] = func15(0.01)
warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 384 column 13
    limit at line 92 column 5
    func15 at line 16 column 9

text_result =
           ⎛101⎞
    100⋅log⎜───⎟
           ⎝100⎠

numeric_result = 0.9950
令\(x = 0\)，计算结果如下
>> [text_result, numeric_result] = func15(0)
text_result =
    0

numeric_result = 0
令\(x \lt 0 ( x = -0.01)\)，计算结果如下
>> [text_result, numeric_result] = func15(-0.01)
warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
    double_to_sym_heuristic at line 50 column 7
    sym at line 384 column 13
    limit at line 92 column 5
    func15 at line 16 column 9

text_result =
    -30⋅√11 + 10⋅√101

numeric_result = 1.0000
因为\( f(0+0) = f(0-0) \ne f(0) \)，所以\( x=0 \)为\( f(x) \)的可去间断点
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