Alternative way to solve a rational limit in $\Bbb C$ when factorization and denominator conjugation fail

Question

Problem

Show that $$\lim_{z\to w}{F(z)} = 6+8i$$ where $F(z)=\dfrac{z^2+7-24i}{z-3-4i}$ and $w=3+4i$ and $z=x+iy$.

Failed Approaches

I called $n=z^2+7-24i$ and $d=z-3-4i$.

1. Try factoring $n$ into a difference of squares to cancel with $d$; however $\sqrt{7-24i}=4-3i$.

2. Try evaluating $F(w)$ by direct substitution, which gives $d=0$ and is undefined.

3. Try evaluating $F(w)$ by considering $\dfrac{n\overline d}{d\overline d}$; however $\overline d = (x-3)-i(y-4)$ evaluates to $0$ at $z=3+4i$ which would imply that $F(w)=0.$

Question

The fact that the question stem suggested to parametrize $z$ into real and imaginary parts suggests to me that I should invoke some theorem about $f(z)\to L$ if $\renewcommand{\Re}{\operatorname{Re}} \Re f(z) \to \Re L$ and $\renewcommand{\Im}{\operatorname{Im}} \Im f(z) \to \Im L$.

What other approaches are there to a problem like this? What step might be next with this particular limit?

Answer 1

Hint: Use L'Hospital's rule, then $$\lim_{z\to3+4i}\dfrac{z^2+7-24i}{z-3-4i}=\lim_{z\to3+4i}2z=6+8i$$

Answer 2

$$F(z) = \frac{z^2+7-24i}{z-3-4i} = \frac{(z-3-4i)(z+3+4i)}{z-3-4i} = z+3+4i $$ So that $$\lim_{z\to 3+4i} F(z) = \lim_{z\to 3+4i} (z+3+4i) = 6+8i $$

Answer 3

A problem that gives you the answer! Instead of telling you to find $$\lim_{z\to3+4i}\frac{z^2+7-24i}{z-(3+4i)}$$ it even tells you what that answer should be! If the answer is going to be $6+8i$ then the numerator had better be $$(z-(3+4i))(z-(3+4i)+6+8i).$$ Is it? Does that equal $z^2+7-24i$?