Questions: Limits and Continuity of Complex Functions

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

Consider f(z) = Re(z)/|z| for z ≠ 0 and f(0) = 0. What happens as z → 0?

AThe limit is 0, because both the numerator and denominator approach 0
BThe limit is 1, because Re(z) ≈ |z| for real z
CThe limit does not exist, because the value along the real axis differs from the value along the imaginary axis
DThe limit exists and equals 1/2 by averaging the real and imaginary approach directions
Question 2 Multiple Choice

Which statement correctly describes the component criterion for limits of complex functions?

Alim_{z→z₀} f(z) = L if and only if |f(z) - L| < ε whenever z is within δ of z₀, for the single approach along the real axis
Blim_{z→z₀} f(z) = L if and only if lim_{(x,y)→(x₀,y₀)} u(x,y) = Re(L) and lim_{(x,y)→(x₀,y₀)} v(x,y) = Im(L) as real multivariable limits
Clim_{z→z₀} f(z) = L if and only if the real and imaginary parts of f are bounded near z₀
Dlim_{z→z₀} f(z) = L if and only if f has no poles in a neighborhood of z₀
Question 3 True / False

If lim_{z→z₀} f(z) = L along the real axis and lim_{z→z₀} f(z) = L along the imaginary axis (both giving the same value L), then lim_{z→z₀} f(z) = L.

TTrue
FFalse
Question 4 True / False

A complex function f(z) = u(x,y) + iv(x,y) is continuous at z₀ = x₀ + iy₀ if and only if both u and v are continuous as real functions from ℝ² to ℝ at (x₀, y₀).

TTrue
FFalse
Question 5 Short Answer

Why is approaching a point z₀ in the complex plane fundamentally more restrictive than approaching x₀ on the real line, and what does this foreshadow about complex differentiability?

Think about your answer, then reveal below.