Annette Arroyo

2020-10-28

Let ${a}_{n}\to 0$, and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined):
$\underset{n\to \mathrm{\infty }}{lim}\frac{1+2{a}_{n}}{1+3{a}_{n}-4\left({a}_{n}{\right)}^{2}}$

To compute, $\underset{n\to \mathrm{\infty }}{lim}\frac{1+2{a}_{n}}{1+3{a}_{n}-4\left({a}_{n}{\right)}^{2}}$
Given that, $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0$
Algebraic limit theorem: Let ${a}_{n}$ and ${b}_{n}$ be sequences of real numbers such that
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=a$ and $\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=b$, Then following statements hold.
a) $\underset{n\to \mathrm{\infty }}{lim}c\cdot {a}_{n}=c\cdot a$
b) $\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}+{b}_{n}\right)=a+b$
c) $\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}\cdot {b}_{n}\right)=a\cdot b$
d) If ${b}_{n}\ne 0$, for all n, then $\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=\frac{a}{b}$
Using the algebraic limit theorem,
$\underset{n\to \mathrm{\infty }}{lim}\frac{1+2{a}_{n}}{\left(1+3\cdot {a}_{n}-4\left({a}_{n}{\right)}^{2}}$
$=\frac{\underset{n\to \mathrm{\infty }}{lim}1+\underset{n\to \mathrm{\infty }}{lim}2{a}_{n}}{\underset{n\to \mathrm{\infty }}{lim}1+\underset{n->\mathrm{\infty }}{lim}3{a}_{n}-\underset{n\to \mathrm{\infty }}{lim}4\left({a}_{n}^{2}}$
$=\frac{\underset{n\to \mathrm{\infty }}{lim}1+2\cdot \underset{n\to \mathrm{\infty }}{lim}{a}_{n}}{\underset{n\to \mathrm{\infty }}{lim}1+\underset{n\to \mathrm{\infty }}{lim}{a}_{n}-4\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}{\right)}^{2}}$
Using given $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0$
$\underset{n\to \mathrm{\infty }}{lim}\frac{1+2{a}_{n}}{1+3{a}_{n}-4\left({a}_{n}{\right)}^{2}}$
$=\frac{1+2\cdot \left(0\right)}{1+3\cdot \left(0\right)-4\cdot \left(0{\right)}^{2}}$
$=\frac{1}{1}$
$=1$
Therefore, $\underset{n\to \mathrm{\infty }}{lim}\frac{1+2{a}_{n}}{1+3{a}_{n}-4\left({a}_{n}{\right)}^{2}}=1$

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