# Let a_nrightarrow0, and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined): lim_{nrightarrowinfty}frac{1+2a_n}{1+3a_n-4(a_n)^2}

Limits and continuity
Let $$a_n\rightarrow0$$, and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined):
$$\lim_{n\rightarrow\infty}\frac{1+2a_n}{1+3a_n-4(a_n)^2}$$

2020-10-29
To compute, $$\lim_{n\rightarrow\infty}\frac{1+2a_n}{1+3a_n-4(a_n)^2}$$
Given that, $$\lim_{n\rightarrow\infty}a_n=0$$
Algebraic limit theorem: Let $$a_n$$ and $$b_n$$ be sequences of real numbers such that
$$\lim_{n\rightarrow\infty}a_n=a$$ and $$\lim_{n\rightarrow\infty}b_n=b$$, Then following statements hold.
a) $$\lim_{n\rightarrow\infty}c\cdot a_n=c\cdot a$$
b) $$\lim_{n\rightarrow\infty}(a_n+b_n)=a+b$$
c) $$\lim_{n\rightarrow\infty}(a_n\cdot b_n)=a\cdot b$$
d) If $$b_n\neq0$$, for all n, then $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\frac{a}{b}$$
Using the algebraic limit theorem,
$$\lim_{n\rightarrow\infty}\frac{1+2a_n}{(1+3\cdot a_n-4(a_n)^2}$$
$$=\frac{\lim_{n\rightarrow\infty}1+\lim_{n\rightarrow\infty}2a_n}{\lim_{n\rightarrow\infty}1+\lim_{n->\infty}3a_n-\lim_{n\rightarrow\infty}4(a_n^2}$$
$$=\frac{\lim_{n\rightarrow\infty}1+2\cdot\lim_{n\rightarrow\infty}a_n}{\lim_{n\rightarrow\infty}1+\lim_{n\rightarrow\infty}a_n-4\lim_{n \rightarrow\infty}(a_n)^2}$$
Using given $$\lim_{n\rightarrow\infty}a_n=0$$
$$\lim_{n\rightarrow\infty}\frac{1+2a_n}{1+3a_n-4(a_n)^2}$$
$$=\frac{1+2\cdot(0)}{1+3\cdot(0)-4\cdot(0)^2}$$
$$=\frac{1}{1}$$
$$=1$$
Therefore, $$\lim_{n\rightarrow\infty}\frac{1+2a_n}{1+3a_n-4(a_n)^2}=1$$