Mark
Use the definition of limit to prove that
\lim_{z \to i}\frac{z^2+1}{z-i} = 2i.
An archive the questions from Mark's Fall 2018 Complex Variables Course.
A limit proof
Great_Big
Let’s try this.
Claim: \lim_{z \to i} \frac{z^2 +1}{z-i}=2i.
Proof:
First notice that \lim_{z \to i} \frac{z^2 +1}{z-i}=2i \Leftrightarrow \lim_{z \to i} \frac{(z+i)(z-i)}{z-i} = 2i \Leftrightarrow \lim_{z \to i} (z+i) =2i.
Now we need to show (\forall \varepsilon>0)(\exists\delta>0)(0<|z-i|<\delta \Rightarrow |(z+i)-2i|< \varepsilon).
So let \varepsilon > 0 and let \delta = \varepsilon. So then 0<|z-i|<\delta implies that
|(z+i)-2i|=|z-i|<\delta= \varepsilon.
Thus, we have shown that which was to be demonstrated.
jgomez28314
\lim_{z\rightarrow i}\frac{z^2+1}{z-1} = 2i
Proof:
First notice that
lim_{z\rightarrow i} \frac{z^2+1}{z-1} = 2i \Leftrightarrow lim_{z\rightarrow i} \frac{(z+i)(z-i)}{z-1} \Leftrightarrow lim_{z\rightarrow i} (z+1) = 2i
now we need to show
(\forall \epsilon > 0)(\exists \delta > 0)(0<|z-1| < \delta \implies |(z+i)-2i| < \epsilon).
So let,
\epsilon > 0 \hspace{2mm} and \hspace{2mm} let \hspace{2mm} \delta = \epsilon.) So \hspace{2mm} then \hspace{2mm}, (0 < |z-i| < \delta \hspace{2mm}),
implies \hspace{2mm} that \hspace{2mm} (|(z+i) - 2i| = |z-i| < \delta = \epsilon. \nabla