Let S ⊆ C S \subseteq \mathbb{C} S ⊆ C f : S → C f: S \to \mathbb{C} f : S → C function , and let z 0 z_0 z 0 limit point of S S S z 0 z_0 z 0 S S S z 0 z_0 z 0 S S S ω 0 ∈ C \omega_0 \in \mathbb{C} ω 0 ∈ C
lim z → z 0 f ( z ) = ω 0 \lim_{z \to z_0} f(z) = \omega_0 z → z 0 lim f ( z ) = ω 0 iff , for each ϵ > 0 \epsilon > 0 ϵ > 0 δ > 0 \delta > 0 δ > 0 z ∈ S z \in S z ∈ S
0 < ∣ z − z 0 ∣ < δ ⟹ ∣ f ( z ) − ω 0 ∣ < ϵ 0 < |z-z_0| < \delta \implies |f(z)-\omega_0| < \epsilon 0 < ∣ z − z 0 ∣ < δ ⟹ ∣ f ( z ) − ω 0 ∣ < ϵ f ( z ) f(z) f ( z ) ω 0 \omega_0 ω 0 z z z z 0 z_0 z 0
Uniqueness of Limits
If a limit of f : S → C f: S \to \mathbb{C} f : S → C S S S
Limits via Real and Imaginary Parts
Let f ( z ) = u ( x , y ) + i v ( x , y ) f(z) = u(x,y)+iv(x,y) f ( z ) = u ( x , y ) + i v ( x , y ) z 0 = x 0 + i y 0 z_0 = x_0+iy_0 z 0 = x 0 + i y 0 ω 0 = u 0 + i v 0 \omega_0 = u_0+iv_0 ω 0 = u 0 + i v 0
lim z → z 0 f ( z ) = ω 0 \lim_{z \to z_0} f(z) = \omega_0 z → z 0 lim f ( z ) = ω 0 iff
lim ( x , y ) → ( x 0 , y 0 ) u ( x , y ) = u 0 and lim ( x , y ) → ( x 0 , y 0 ) v ( x , y ) = v 0 \lim_{(x,y) \to (x_0,y_0)} u(x,y) = u_0 \quad \text{and} \quad \lim_{(x,y) \to (x_0,y_0)} v(x,y) = v_0 ( x , y ) → ( x 0 , y 0 ) lim u ( x , y ) = u 0 and ( x , y ) → ( x 0 , y 0 ) lim v ( x , y ) = v 0 Limit Laws
Suppose lim z → z 0 f ( z ) = ω 0 \displaystyle\lim_{z \to z_0} f(z) = \omega_0 z → z 0 lim f ( z ) = ω 0 lim z → z 0 g ( z ) = Ω 0 \displaystyle\lim_{z \to z_0} g(z) = \Omega_0 z → z 0 lim g ( z ) = Ω 0
Sum Law
lim z → z 0 ( f ( z ) + g ( z ) ) = ω 0 + Ω 0 \displaystyle\lim_{z \to z_0} (f(z)+g(z)) = \omega_0+\Omega_0 z → z 0 lim ( f ( z ) + g ( z )) = ω 0 + Ω 0 Proof Hint
Fix ϵ > 0 \epsilon > 0 ϵ > 0
Choose δ 1 \delta_1 δ 1 0 < ∣ z − z 0 ∣ < δ 1 ⟹ ∣ f ( z ) − ω 0 ∣ < ϵ / 2 0 < |z-z_0| < \delta_1 \implies |f(z)-\omega_0| < \epsilon/2 0 < ∣ z − z 0 ∣ < δ 1 ⟹ ∣ f ( z ) − ω 0 ∣ < ϵ /2
Choose δ 2 \delta_2 δ 2 0 < ∣ z − z 0 ∣ < δ 2 ⟹ ∣ g ( z ) − Ω 0 ∣ < ϵ / 2 0 < |z-z_0| < \delta_2 \implies |g(z)-\Omega_0| < \epsilon/2 0 < ∣ z − z 0 ∣ < δ 2 ⟹ ∣ g ( z ) − Ω 0 ∣ < ϵ /2
Let δ = min ( δ 1 , δ 2 ) \delta = \min(\delta_1, \delta_2) δ = min ( δ 1 , δ 2 )
Rewrite ∣ f ( z ) + g ( z ) − ( ω 0 + Ω 0 ) ∣ |f(z)+g(z)-(\omega_0+\Omega_0)| ∣ f ( z ) + g ( z ) − ( ω 0 + Ω 0 ) ∣ ∣ ( f ( z ) − ω 0 ) + ( g ( z ) − Ω 0 ) ∣ |(f(z)-\omega_0)+(g(z)-\Omega_0)| ∣ ( f ( z ) − ω 0 ) + ( g ( z ) − Ω 0 ) ∣
Apply the triangle inequality
Product Law
lim z → z 0 ( f ( z ) g ( z ) ) = ω 0 Ω 0 \displaystyle\lim_{z \to z_0} (f(z)g(z)) = \omega_0\Omega_0 z → z 0 lim ( f ( z ) g ( z )) = ω 0 Ω 0 Proof Hint
Rewrite f ( z ) g ( z ) − ω 0 Ω 0 f(z)g(z)-\omega_0\Omega_0 f ( z ) g ( z ) − ω 0 Ω 0 ( f ( z ) − ω 0 ) g ( z ) + ω 0 ( g ( z ) − Ω 0 ) (f(z)-\omega_0)g(z)+\omega_0(g(z)-\Omega_0) ( f ( z ) − ω 0 ) g ( z ) + ω 0 ( g ( z ) − Ω 0 )
Apply the definition of the limit of g g g ϵ = 1 \epsilon = 1 ϵ = 1 ∣ g ( z ) ∣ < ∣ Ω 0 ∣ + 1 |g(z)| < |\Omega_0|+1 ∣ g ( z ) ∣ < ∣ Ω 0 ∣ + 1 z 0 z_0 z 0
Choose δ 1 \delta_1 δ 1 ∣ f ( z ) − ω 0 ∣ < ϵ 2 ( ∣ Ω 0 ∣ + 1 ) |f(z)-\omega_0| < \dfrac{\epsilon}{2(|\Omega_0|+1)} ∣ f ( z ) − ω 0 ∣ < 2 ( ∣ Ω 0 ∣ + 1 ) ϵ
Choose δ 2 \delta_2 δ 2 ∣ g ( z ) − Ω 0 ∣ < ϵ 2 ( ∣ ω 0 ∣ + 1 ) |g(z)-\Omega_0| < \dfrac{\epsilon}{2(|\omega_0|+1)} ∣ g ( z ) − Ω 0 ∣ < 2 ( ∣ ω 0 ∣ + 1 ) ϵ
Let δ \delta δ
Apply the triangle inequality
Quotient Law
lim z → z 0 f ( z ) g ( z ) = ω 0 Ω 0 , provided Ω 0 ≠ 0 \displaystyle\lim_{z \to z_0} \frac{f(z)}{g(z)} = \frac{\omega_0}{\Omega_0}, \quad \text{provided} \quad \Omega_0 \ne 0 z → z 0 lim g ( z ) f ( z ) = Ω 0 ω 0 , provided Ω 0 = 0 Proof Hint
Prove lim z → z 0 1 g ( z ) = 1 Ω 0 \displaystyle\lim_{z \to z_0} \frac{1}{g(z)} = \frac{1}{\Omega_0} z → z 0 lim g ( z ) 1 = Ω 0 1 f ( z ) f(z) f ( z )
Apply the definition of the limit of g g g ϵ = ∣ Ω 0 ∣ / 2 \epsilon = |\Omega_0|/2 ϵ = ∣ Ω 0 ∣/2 ∣ g ( z ) ∣ > ∣ Ω 0 ∣ / 2 |g(z)| > |\Omega_0|/2 ∣ g ( z ) ∣ > ∣ Ω 0 ∣/2 z 0 z_0 z 0
Rewrite ∣ 1 g ( z ) − 1 Ω 0 ∣ \left|\dfrac{1}{g(z)}-\dfrac{1}{\Omega_0}\right| g ( z ) 1 − Ω 0 1 ∣ g ( z ) − Ω 0 ∣ ∣ g ( z ) ∣ ∣ Ω 0 ∣ \dfrac{|g(z)-\Omega_0|}{|g(z)||\Omega_0|} ∣ g ( z ) ∣∣ Ω 0 ∣ ∣ g ( z ) − Ω 0 ∣
Bound it by 2 ∣ g ( z ) − Ω 0 ∣ ∣ Ω 0 ∣ 2 \dfrac{2|g(z)-\Omega_0|}{|\Omega_0|^2} ∣ Ω 0 ∣ 2 2∣ g ( z ) − Ω 0 ∣
Choose δ \delta δ ∣ g ( z ) − Ω 0 ∣ < ϵ ∣ Ω 0 ∣ 2 2 |g(z)-\Omega_0| < \dfrac{\epsilon|\Omega_0|^2}{2} ∣ g ( z ) − Ω 0 ∣ < 2 ϵ ∣ Ω 0 ∣ 2
Limit of a Composition
Let f : B → C f: B \to \mathbb{C} f : B → C g : A → B g: A \to B g : A → B
lim ω → ω 0 g ( ω ) = z 0 and lim z → z 0 f ( z ) = L \lim_{\omega \to \omega_0} g(\omega) = z_0 \quad \text{and} \quad \lim_{z \to z_0} f(z) = L ω → ω 0 lim g ( ω ) = z 0 and z → z 0 lim f ( z ) = L and g ( ω ) ≠ z 0 g(\omega) \ne z_0 g ( ω ) = z 0 ω \omega ω ω 0 \omega_0 ω 0 ω ≠ ω 0 \omega \ne \omega_0 ω = ω 0
lim ω → ω 0 f ( g ( ω ) ) = L \lim_{\omega \to \omega_0} f(g(\omega)) = L ω → ω 0 lim f ( g ( ω )) = L Proof Hint
Fix ϵ > 0 \epsilon > 0 ϵ > 0
Choose δ 1 \delta_1 δ 1 0 < ∣ z − z 0 ∣ < δ 1 ⟹ ∣ f ( z ) − L ∣ < ϵ 0 < |z-z_0| < \delta_1 \implies |f(z)-L| < \epsilon 0 < ∣ z − z 0 ∣ < δ 1 ⟹ ∣ f ( z ) − L ∣ < ϵ
Choose δ 2 \delta_2 δ 2 0 < ∣ ω − ω 0 ∣ < δ 2 ⟹ ∣ g ( ω ) − z 0 ∣ < δ 1 0 < |\omega-\omega_0| < \delta_2 \implies |g(\omega)-z_0| < \delta_1 0 < ∣ ω − ω 0 ∣ < δ 2 ⟹ ∣ g ( ω ) − z 0 ∣ < δ 1
Choose δ 3 \delta_3 δ 3 0 < ∣ ω − ω 0 ∣ < δ 3 ⟹ g ( ω ) ≠ z 0 0 < |\omega-\omega_0| < \delta_3 \implies g(\omega) \ne z_0 0 < ∣ ω − ω 0 ∣ < δ 3 ⟹ g ( ω ) = z 0
Let δ = min ( δ 2 , δ 3 ) \delta = \min(\delta_2, \delta_3) δ = min ( δ 2 , δ 3 )
Conclude 0 < ∣ g ( ω ) − z 0 ∣ < δ 1 0 < |g(\omega)-z_0| < \delta_1 0 < ∣ g ( ω ) − z 0 ∣ < δ 1 ∣ f ( g ( ω ) ) − L ∣ < ϵ |f(g(\omega))-L| < \epsilon ∣ f ( g ( ω )) − L ∣ < ϵ
Limits Involving Infinity
Let z 0 , ω 0 ∈ C z_0, \omega_0 \in \mathbb{C} z 0 , ω 0 ∈ C
Infinite Limit at a Point
lim z → z 0 f ( z ) = ∞ ⟺ lim z → z 0 1 f ( z ) = 0 \lim_{z \to z_0} f(z) = \infty \iff \lim_{z \to z_0} \frac{1}{f(z)} = 0 z → z 0 lim f ( z ) = ∞ ⟺ z → z 0 lim f ( z ) 1 = 0 where f ( z ) ≠ 0 f(z) \ne 0 f ( z ) = 0 z z z z 0 z_0 z 0 z ≠ z 0 z \ne z_0 z = z 0
Proof Hint
Start with the definition lim z → z 0 f ( z ) = ∞ \displaystyle\lim_{z \to z_0} f(z) = \infty z → z 0 lim f ( z ) = ∞ iff ∀ M > 0 , ∃ δ > 0 \forall M > 0, \exists \delta > 0 ∀ M > 0 , ∃ δ > 0 0 < ∣ z − z 0 ∣ < δ ⟹ ∣ f ( z ) ∣ > M 0 < |z-z_0| < \delta \implies |f(z)| > M 0 < ∣ z − z 0 ∣ < δ ⟹ ∣ f ( z ) ∣ > M
Set M = 1 / ϵ M = 1/\epsilon M = 1/ ϵ
Rewrite ∣ f ( z ) ∣ > 1 / ϵ |f(z)| > 1/\epsilon ∣ f ( z ) ∣ > 1/ ϵ ∣ 1 f ( z ) − 0 ∣ < ϵ \left|\dfrac{1}{f(z)}-0\right| < \epsilon f ( z ) 1 − 0 < ϵ
Reverse the steps for the other direction
Finite Limit at Infinity
lim z → ∞ f ( z ) = ω 0 ⟺ lim z → 0 f ( 1 z ) = ω 0 \lim_{z \to \infty} f(z) = \omega_0 \iff \lim_{z \to 0} f\left(\frac{1}{z}\right) = \omega_0 z → ∞ lim f ( z ) = ω 0 ⟺ z → 0 lim f ( z 1 ) = ω 0 Proof Hint
Start with the definition lim z → ∞ f ( z ) = ω 0 \displaystyle\lim_{z \to \infty} f(z) = \omega_0 z → ∞ lim f ( z ) = ω 0 iff ∀ ϵ > 0 , ∃ R > 0 \forall \epsilon > 0, \exists R > 0 ∀ ϵ > 0 , ∃ R > 0 ∣ z ∣ > R ⟹ ∣ f ( z ) − ω 0 ∣ < ϵ |z| > R \implies |f(z)-\omega_0| < \epsilon ∣ z ∣ > R ⟹ ∣ f ( z ) − ω 0 ∣ < ϵ
Set δ = 1 / R \delta = 1/R δ = 1/ R
Substitute z ↦ 1 / z z \mapsto 1/z z ↦ 1/ z 0 < ∣ z ∣ < δ 0 < |z| < \delta 0 < ∣ z ∣ < δ ∣ 1 / z ∣ > R |1/z| > R ∣1/ z ∣ > R
Reverse the steps for the other direction
Infinite Limit at Infinity
lim z → ∞ f ( z ) = ∞ \displaystyle\lim_{z \to \infty} f(z) = \infty z → ∞ lim f ( z ) = ∞ iff ∀ M > 0 , ∃ R > 0 \forall M > 0, \exists R > 0 ∀ M > 0 , ∃ R > 0 ∣ z ∣ > R ⟹ ∣ f ( z ) ∣ > M |z| > R \implies |f(z)| > M ∣ z ∣ > R ⟹ ∣ f ( z ) ∣ > M
lim z → ∞ f ( z ) = ∞ ⟺ lim z → 0 1 f ( 1 / z ) = 0 \lim_{z \to \infty} f(z) = \infty \iff \lim_{z \to 0} \frac{1}{f(1/z)} = 0 z → ∞ lim f ( z ) = ∞ ⟺ z → 0 lim f ( 1/ z ) 1 = 0 Proof Hint
Set M = 1 / ϵ M = 1/\epsilon M = 1/ ϵ δ = 1 / R \delta = 1/R δ = 1/ R
Substitute z ↦ 1 / z z \mapsto 1/z z ↦ 1/ z
Rewrite ∣ f ( 1 / z ) ∣ > 1 / ϵ |f(1/z)| > 1/\epsilon ∣ f ( 1/ z ) ∣ > 1/ ϵ ∣ 1 f ( 1 / z ) ∣ < ϵ \left|\dfrac{1}{f(1/z)}\right| < \epsilon f ( 1/ z ) 1 < ϵ
Reverse the steps for the other direction
