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Is $g(z + \Delta z).g(z) = g(z)^2$ ?

The full expression is a $\lim_{\Delta z \to 0} \frac {A}{G} $, where $G$ is the left-hand side of the above expression.

It's a question about a solution to a problem that I'm going through, and I can't post it because I don't have enough reps. But the gist of it is that the above expression is taken to be true and thus he factors a $g(z)^2$ out of the limits in the denominator.

I'd appreciate it if somebody explains how this is possible. Thanks!

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if $g(z) $ is continuous at $z$. – Aang Jul 13 '12 at 9:13
And if the statement is $\lim\limits_{\Delta z\to0}g(z+\Delta z)\cdot g(z)=g(z)^2$. – Did Jul 13 '12 at 9:22
So, as a general rule, when evaluating limits I guess I can plug in the value of the limit into the appropriate place and evaluate the expression? (It's been a while since I did these). – Joebevo Jul 13 '12 at 10:10
up vote 2 down vote accepted

Let us see a general answer to your particular question. Let $g$ be a function, and assume that $g$ is continuous at $z_0$. Then $$ \lim_{\Delta z \to 0} g(z_0+\Delta z) = g(z_0), \tag{1} $$ and hence $$ \lim_{\Delta z \to 0} g(z_0+\Delta z)g(z_0) = g(z_0)\cdot g(z_0)=g(z_0)^2. $$ Since (1) is the definition of continuity, the answer to your questoin is affirmative provided that $g$ is continuous. If $g$ is not supposed to be continuous, then then answer might be negative: you should be able to use any reasonable kind of discontinuous counterexample to check this.

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