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## Homework Statement

Prove that $$\frac{x_n^2 - e}{x_n} \rightarrow 1-e$$ as ##n \rightarrow \infty##, provided ##x_n \rightarrow 1## as n ##\rightarrow \infty##.

## The Attempt at a Solution

The above holds if ##\,\forall\, \epsilon > 0 \,\exists \, N\, \in\, \mathbb{N}## such that if n ##\geq N## then ##|\frac{x_n^2 - e}{x_n} - (1-e)| < \epsilon##. Work with the expression in the mod signs and rewrite as: $$|x_n - \frac{e}{x_n} - 1 +e| \leq |x_n - 1| + |e-\frac{e}{x_n}|$$using the ##\Delta## inequality. The first term is ##< \epsilon## and the additional term on the RHS is positive, so can I just stop here?

If not, I proceeded using the fact that ##|x_n-1| < \epsilon## and established that ##|-\frac{1}{x_n}| < \frac{1}{1-\epsilon}\,\,(1)## This gives an upper bound entiirely in terms of ##\epsilon##. Since the statement should hold for all ##\epsilon##,then I would usually finsih here. However, I notice that if ##\epsilon =1## then I end up with nonsense. Is there something else I can try?

Many thanks.