Suppose that $x_n$ is a sequence of real numbers that converges to $1$ as $n\to\infty$. Using the definition, prove that following limit exists. $$\frac{x^2_n − e}{x_n}→ 1 − e \quad\text{ as }\quad n \to\infty.$$
Here is my attempt, Is there anything wrong here?
By hypothesis, given $\epsilon>0$ there is an $N_1\in\mathbb{N}$ such that $n\geq N _1$ implies $|x_n-1|<\frac{\epsilon}{5}$.
$$|\frac{x_n^2-e}{x_n}-(1-e)|=|x_n-\frac{e}{x_n}-1+e|=|(x_n-1)+e(\frac{x_n-1}{x_n})|$$
since $e<3$ and $\epsilon=\frac{5}{4}$ choose $N_2$ sutch that $n≥N_2$ implies $|x_n-1|<\frac{1}{4}$, (i.e $x_n>\frac{3}{4})$ set $N=\max(N_1,N_2)$
$$|\frac{x_n^2-e}{x_n}-(1-e)|=|x_n-1||1+\frac{e}{x_n}|<|x_n-1|(1+\frac{3}{x_n})<5|x_n-1|<\epsilon$$
for all $n≥N$.
Thank you!!!
That is not correct. You cannot decide that $\varepsilon=\frac54$; it can be any number greater than $0$.
You can do it as follows. Take $\varepsilon>0$. Now, take $N_1\in\Bbb N$ such that $n\geqslant N_1\implies|x_n-1|<\frac\varepsilon2$. Then, take $N_2\in\Bbb N$ such that $n\geqslant N_2\implies|x_n-1|<\frac12$ and note that this implies that $|x_n|>\frac12$. Then, take $N_3\in\Bbb N$ such that $n\geqslant N_3\implies|x_n-1|<\frac1{4e}\varepsilon$. Now, if $N=\max\{N_1,N_2,N_3\}$, then\begin{align}n\geqslant N\implies\left|\frac{x_n^{\,2}-e}{x_n}-(1-e)\right|&=\left|x_n-1+e\left(\frac{x_n-1}{x_n}\right)\right|\\&\leqslant|x_n-1|+2e|x_n-1|\text{ (because $\frac1{|x_n|}<2$)}\\&\leqslant\frac\varepsilon2+\frac\varepsilon2\\&=\varepsilon.\end{align}