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Question Number 60597 by Tinkutara last updated on 22/May/19

Commented by Tinkutara last updated on 22/May/19

Should a be the answer or not?

Commented by ajfour last updated on 22/May/19

i believe, no .  l depends of course on n(z).  Further does lim_(z→0) n(z)=n_1   and lim_(z→d) n(z)=n_2   or not?

$$\mathrm{i}\:\mathrm{believe},\:\mathrm{no}\:. \\ $$$${l}\:\mathrm{depends}\:\mathrm{of}\:\mathrm{course}\:\mathrm{on}\:\mathrm{n}\left(\mathrm{z}\right). \\ $$$$\mathrm{Further}\:\mathrm{does}\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}n}\left(\mathrm{z}\right)=\mathrm{n}_{\mathrm{1}} \\ $$$$\mathrm{and}\:\underset{\mathrm{z}\rightarrow\mathrm{d}} {\mathrm{lim}n}\left(\mathrm{z}\right)=\mathrm{n}_{\mathrm{2}} \:\:\mathrm{or}\:\mathrm{not}? \\ $$

Commented by Tinkutara last updated on 22/May/19

Sorry Sir it's a misprint in book, got it now.

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