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}

Question
Limits and continuity
asked 2020-10-28
Let \(a_n\rightarrow0\), and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined):
\(\lim_{n\rightarrow\infty}\frac{1+2a_n}{1+3a_n-4(a_n)^2}\)

Answers (1)

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, \(\lim_{n\rightarrow\infty}\frac{1+2a_n}{1+3a_n-4(a_n)^2}=1\)
0

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