- #1

- 219

- 0

## Main Question or Discussion Point

Suppose f(z) is analytic in a region R including the point z

this is actually a lemma my book proves first before actually proving L'Hospital's rule. I understood how they used the lemma to prove the rule but i don't really understand the logic in proving this lemma. my book did:

Let [f(z) - f(z

Then, since f(z) is analytic at z

lim (z ~> z

i don't understand how f(z) = f(z

_{0}. Prove that f(z) = f(z_{0}) + f'(z_{0})(z-z_{0}) + η(z-z_{0}) where η ~> 0 as z ~> z_{0}.this is actually a lemma my book proves first before actually proving L'Hospital's rule. I understood how they used the lemma to prove the rule but i don't really understand the logic in proving this lemma. my book did:

Let [f(z) - f(z

_{0})]/(z-z_{0}) - f'(z_{0}) = η so that f(z) = f(z_{0}) + f'(z_{0})(z - z_{0}) = η(z-z_{0}).Then, since f(z) is analytic at z

_{0}, we have as required:lim (z ~> z

_{0}) of η = lim (z ~> z_{0}) of [f(z) - f(z_{0})]/(z-z_{0}) - f'(z_{0}) = f'(z_{0}) - f'(z_{0}) = 0.i don't understand how f(z) = f(z

_{0}) + f'(z_{0})(z - z_{0}) = η(z-z_{0}). shouldn't it be f(z) = η(z-z_{0}) + f'(z_{0})(z - z_{0}) + f(z_{0}) since they let [f(z) - f(z_{0})]/(z-z_{0}) - f'(z_{0}) = η?