Find $\lim_{z\to 1+i}\left(\frac{z+2-i}{3}\right)^\frac{z-i}{z-1-i}$

Question

I am trying to find the limit

$$ \lim_{z\to 1+i}\left(\frac{z+2-i}{3}\right)^\frac{z-i}{z-1-i} $$

If we plug in $1+i$ for $z$, we get the following

$$ \lim_{z\to 1+i}\left(\frac{z+2-i}{3}\right)^\frac{z-i}{z-1-i}=1^\infty$$

Can we say that this limit equals to $1$?

Answer 1

Letting $\omega=z-1-i$ converts the limit to \begin{aligned} L & =\lim _{\omega \rightarrow 0}\left(\frac{\omega+3}{3}\right)^{\frac{\omega+1}{\omega}} \\ & =\lim _{\omega \rightarrow 0}\left(1+\frac{\omega}{3}\right)^1\left(1+\frac{\omega}{3}\right)^{\frac{1}{\omega}} \\ & =\left[\lim _{\omega \rightarrow 0}\left(1+\frac{\omega}{3}\right)^{\frac{3}{\omega}}\right]^{\frac{1}{3}} \\ & =e^{\frac{1}{3}} \end{aligned}

Answer 2

While solving this question, one of the standard limits that must be kept in mind is as follows:

$\lim_{x\to0} (1+f(x))^{g(x)} \Longrightarrow e^{\lim_{x\to0} f(x).g(x)}$

On further evaluation:

$\lim_{z\to 1+i}\left(\frac{z+2-i}{3}\right)^\frac{z-i}{z-1-i}\\ \lim_{z\to 1+i}\left(1+\frac{z+2-i}{3}-1\right)^\frac{z-i}{z-1-i}\\ \lim_{z\to 1+i}\left(1+\frac{z-1-i}{3}\right)^\frac{z-i}{z-1-i}\\ e^{\lim_{z\to 1+i}\left(\frac{z-1-i}{3}\right).\left(\frac{z-i}{z-1-i}\right)}\\ e^{\lim_{z\to 1+i}\left(\frac{z-i}{3}\right)}\\ e^{\frac{1+i-i}{3}}\\ e^{\frac{1}{3}}$

So, $e^{\frac{1}{3}}$ is your final answer. Thank you!