помогите решить на паскале 3. Вычислить значение функции tg(x)/x2 на интервале от *-5, 5] c шагом h = 0.1.

Решил!
Вот результат:

f(-5.00) = 0.1352
f(-4.90) = 0.2194
f(-4.80) = 0.4941
f(-4.70) = -3.6538
f(-4.60) = -0.4187
f(-4.50) = -0.2290
f(-4.40) = -0.1599
f(-4.30) = -0.1236
f(-4.20) = -0.1008
f(-4.10) = -0.0847
f(-4.00) = -0.0724
f(-3.90) = -0.0623
f(-3.80) = -0.0536
f(-3.70) = -0.0456
f(-3.60) = -0.0381
f(-3.50) = -0.0306
f(-3.40) = -0.0229
f(-3.30) = -0.0147
f(-3.20) = -0.0057
f(-3.10) = 0.0043
f(-3.00) = 0.0158
f(-2.90) = 0.0293
f(-2.80) = 0.0453
f(-2.70) = 0.0648
f(-2.60) = 0.0890
f(-2.50) = 0.1195
f(-2.40) = 0.1590
f(-2.30) = 0.2116
f(-2.20) = 0.2838
f(-2.10) = 0.3877
f(-2.00) = 0.5463
f(-1.90) = 0.8108
f(-1.80) = 1.3229
f(-1.70) = 2.6632
f(-1.60) = 13.3721
f(-1.50) = -6.2673
f(-1.40) = -2.9581
f(-1.30) = -2.1314
f(-1.20) = -1.7862
f(-1.10) = -1.6238
f(-1.00) = -1.5574
f(-0.90) = -1.5558
f(-0.80) = -1.6088
f(-0.70) = -1.7190
f(-0.60) = -1.9004
f(-0.50) = -2.1852
f(-0.40) = -2.6425
f(-0.30) = -3.4371
f(-0.20) = -5.0678
f(-0.10) = -10.0335
f(0.00) = <на 0 делить нельзя>
f(0.10) = 10.0335
f(0.20) = 5.0678
f(0.30) = 3.4371
f(0.40) = 2.6425
f(0.50) = 2.1852
f(0.60) = 1.9004
f(0.70) = 1.7190
f(0.80) = 1.6088
f(0.90) = 1.5558
f(1.00) = 1.5574
f(1.10) = 1.6238
f(1.20) = 1.7862
f(1.30) = 2.1314
f(1.40) = 2.9581
f(1.50) = 6.2673
f(1.60) = -13.3721
f(1.70) = -2.6632
f(1.80) = -1.3229
f(1.90) = -0.8108
f(2.00) = -0.5463
f(2.10) = -0.3877
f(2.20) = -0.2838
f(2.30) = -0.2116
f(2.40) = -0.1590
f(2.50) = -0.1195
f(2.60) = -0.0890
f(2.70) = -0.0648
f(2.80) = -0.0453
f(2.90) = -0.0293
f(3.00) = -0.0158
f(3.10) = -0.0043
f(3.20) = 0.0057
f(3.30) = 0.0147
f(3.40) = 0.0229
f(3.50) = 0.0306
f(3.60) = 0.0381
f(3.70) = 0.0456
f(3.80) = 0.0536
f(3.90) = 0.0623
f(4.00) = 0.0724
f(4.10) = 0.0847
f(4.20) = 0.1008
f(4.30) = 0.1236
f(4.40) = 0.1599
f(4.50) = 0.2290
f(4.60) = 0.4187
f(4.70) = 3.6538
f(4.80) = -0.4941
f(4.90) = -0.2194
f(5.00) = -0.1352

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