Then, whenever |x - x0| < δ , we have
|1/x - 1/x0| < ε
plt.plot(x, y) plt.title('Plot of f(x) = 1/x') plt.xlabel('x') plt.ylabel('f(x)') plt.grid(True) plt.show() mathematical analysis zorich solutions
|x - x0| < δ .
Using the inequality |1/x - 1/x0| = |x0 - x| / |xx0| ≤ |x0 - x| / x0^2 , we can choose δ = min(x0^2 ε, x0/2) . Then, whenever |x - x0| < δ ,
whenever
|1/x - 1/x0| ≤ |x0 - x| / x0^2 < ε .