# MATH3401

Lecture 13

### Limits at $$\infty$$

Recall that in $$\R$$, $$\ds \lim_{x\ra x_0}f(x)=\infty$$ means:

Given $$M>0$$, there exists $$\delta>0$$ such that $0 < |x - x_0| < \delta \Rightarrow f(x) > M.$

### Limits at $$\infty$$

In $$\C$$, a neighbourhood of $$z_0\in \C$$ is

### Limits at $$\infty$$

A neighbourhood of $$\infty$$ in $$\C$$ has the form $$\{z: |z|>M\}.$$

### Limits at $$\infty$$

"Close to $$\infty$$" $$\iff$$ $$|z|$$ is large $$\iff$$ $$\dfrac{1}{|z|}$$ is small.

Keeping that in mind:

$$(*)$$ $$\ds \lim_{z\ra z_0}f(z) = \infty$$ means $$\ds \lim_{z\ra z_0}\frac{1}{f(z)} = 0.$$

### Limits at $$\infty$$

$$(*)$$ $$\ds \lim_{z\ra \infty}f(z) = w_0$$ means $$\ds \lim_{z\ra 0}f\left(\frac{1}{z}\right) = w_0.$$

$$(*)$$ $$\ds \lim_{z\ra \infty}f(z) = \infty$$ means $$\ds \lim_{z\ra 0}\frac{1}{f\left(\frac{1}{z}\right)} = 0.$$

#### Limits at $$\infty$$

Example 1: Show $$\ds \lim_{z\ra -1}\frac{iz+3}{z+1}= \infty.$$

Let $$f(z)=\ds \frac{iz+3}{z+1}$$. We want to show $$\ds \lim_{z\ra -1}\frac{1}{f\left(z\right)}= 0.$$

LHS $$=\ds \lim_{z\ra -1} \left(\frac{z+1}{iz+3}\right)$$ $$= \frac{\ds \lim_{z\ra -1} (z+1) }{\ds \lim_{z\ra -1} (iz+3) }$$ $$= \ds \frac{0}{3-i} = 0$$ $$=$$ RHS.

(via Lect 12, Theorem 2, part 4)

#### Limits at $$\infty$$

Example 2: Calculate $$\ds \lim_{z\ra \infty}\frac{2z^3-1}{z^2-1}.$$ Let $$g(z)=\ds \frac{2z^3-1}{z^2-1}$$.

Claim $$\ds \lim_{z\ra \infty}g(z)= \infty.$$ So we want to show $$\ds \lim_{z\ra 0}\frac{1}{g\left(\frac{1}{z}\right)} = 0.$$

$$\ds \frac{1}{g\left(\frac{1}{z}\right)}$$ $$=\ds \frac{1}{\frac{2\left(\frac{1}{z}\right)^3-1}{\left(\frac{1}{z}\right)^2-1}}$$ $$= \ds \frac{\frac{1}{z^2}-1}{\frac{2}{z^3}-1}$$ $$= \ds \frac{z-z^3}{2-z^3}$$ $$\ra \ds\frac{0}{2}$$ as $$z\ra 0$$.

### Continuity

Let $$f$$ be a $$\C$$-valued function defined in a neighbourhood of $$z_0$$.

"$$f$$ is continuous at $$z_0$$" means: $$\ds\lim_{z\ra z_0 } f(z) = f(z_0).$$

i. e. given $$\varepsilon >0$$ there exists $$\delta >0$$ such that $|z-z_0|\lt\delta \implies |f(z)-f(z_0)|\lt\varepsilon.$

### Continuity

###### Basic results

1. If $$f: \Omega \ra U$$ and $$g:U\ra W$$ are continuous, so is $$g \circ f : \Omega \ra W$$
$$g$$ composed with $$f$$: $$\left(g\circ f\right)(z) = g\left(\,f(z)\right)$$.

2. If $$f$$ is continuous and nonzero at $$z_0$$ then there exists $$\varepsilon>0$$ such that $$f(z)\neq 0$$ on $$B_{\varepsilon}(z_0)$$. Proof in tutes.

3. $$f: x+iy \mapsto u(x,y) + iv(x,y)$$ is continuous if and only if $$u$$ and $$v$$ are continuous.

4. Obvious analogues of Theorems 1 & 2 from Lecture 12.

### Differentiability in $$\R$$

$$f: \Omega \subseteq \R \ra \R$$, $\lim_{h\ra 0 } \frac{f(x+h)-f(z)}{h}.$

If this exists, defines $$f'(x)$$. Limit in $$\R$$.

### Differentiability in $$\C$$

$$f: \Omega \subseteq \C \ra \C$$, $\lim_{\xi\ra 0 } \frac{f(z_0+\xi)-f(z_0)}{\xi}. \qquad (1)$

If this exists, defines $$f'(z_0)$$.

### Differentiability in $$\C$$

Write $$\Delta z$$ for $$\xi$$.

$$(1)$$ implies $$f'(z_0) = \ds \lim_{\Delta z \ra 0} \frac{f(z_0+\Delta z)- f(z_0)}{\Delta z}.\quad (2)$$

Write $$w = f(z)$$: $$\Delta w = f(z_0 + \Delta) - f(z)$$.

So $$(2)$$ implies $$f'(z_0) = \ds \lim_{\Delta z \ra 0} \frac{\Delta w}{\Delta z} = \frac{dw}{dz}(z_0).\quad (3)$$

Note (1)-(3) are equivalent.