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Could someone verify this proof? Prove $\lim\limits_{n\to\infty} C_n = \lim\limits_{n\to\infty} \dfrac{4n+3}{7n-5} = \dfrac{4}{7}$

Proof: Let $\epsilon > 0$ and take $N = \dfrac{41}{49\epsilon} + 1$ This implies that for all $n > \dfrac{41}{49\epsilon} + 1 \iff n-1 > \dfrac{41}{49\epsilon} \iff \epsilon(n-1) > \dfrac{41}{49} \iff \epsilon > \dfrac{41}{49n - 49} \iff$

$ \iff \epsilon > \dfrac{41}{49n-35} \iff \epsilon > \dfrac{28n+21-28n+20}{49n-35} \iff \dfrac{7(4n+3)-4(7n-5)}{7(7n-5)} \iff \epsilon > \dfrac{4n+3}{7n-5}-\dfrac{4}{7} \iff \epsilon >|\dfrac{4n+3}{7n-5}-\dfrac{4}{7}|$

End of proof

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2 Answers 2

up vote 2 down vote accepted

This is correct I guess, but extremely tedious. I would never actually prove a limit exists this way when we are dealing with quotients of polynomials. The standard thing to do is prove the following theorem:

Theorem. If $\lim\limits_{n \to \infty} a_n = A$ and $\lim\limits_{n \to \infty} b_n = B$ and $c\in \mathbb{R}$ then $$ \lim\limits_{n \to \infty}(c a_n) = c A \text{ as long as $cA \neq 0 \cdot \infty$} $$ $$ \lim\limits_{n \to \infty}(a_n + b_n) = A + B \text{as long as $A + B \neq \infty - \infty$} $$ $$ \lim\limits_{n \to \infty}(a_n b_n) = AB , \text{ as long as $AB \neq 0\cdot \infty$} $$ $$ \lim\limits_{n \to \infty}\frac{a_n}{b_n} = \frac{A}{B}, \text{ as long as $B \neq 0$ and not $B=A=\infty$}. $$

One you have this theorem we can write $$\frac{4n+3}{7n-5} = \frac{4 + \frac{3}{n}}{7 - \frac{5}{n}}$$ for all $n \neq 0$. Applying the quotient of limits part of the previous theorem we find $$ \lim_{n \to \infty}\frac{4n+3}{7n-5} = \lim_{n \to \infty}\frac{4 + \frac{3}{n}}{7 - \frac{5}{n}} = \frac{\lim\limits_{n \to \infty}(4+\frac{3}{n})}{\lim\limits_{n \to \infty}(7 - \frac{5}{n})} = \frac{4+0}{7+0} = \frac{4}{7}. $$

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I would work forwards, not backwards starting with magic constants.

We want to show that $lim_{n \to \infty} \frac{4n+3}{7n-5} = \frac{4}{7}$.

$\frac{4n+3}{7n-5} - \frac{4}{7} = \frac{7(4n+3)-4(7n-5)}{7(7n-5)} = \frac{28n+21-28n+20}{49n-35} = \frac{41}{49n-35} $.

To make this to be less than $c$ for any real $c$, choose $n$ such that $\frac{41}{49n-35} < c$ or $41 < 49cn-35c$ or $n > \frac{41+35c}{49n}$.

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