Exercise 2.3.4

Question

Let $\left(a_{n}\right) \rightarrow 0$, and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined):
  \begin{enumerate}[label=(\alph*)] 
  \item $\lim \left(\frac{1+2 a_{n}}{1+3 a_{n}-4 a_{n}^{2}}\right)$
  \item $\lim \left(\frac{\left(a_{n}+2\right)^{2}-4}{a_{n}}\right)$
  \item $\lim \left(\frac{\frac{2}{a_{n}}+3}{\frac{1}{a_{n}}+5}\right)$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Apply the ALT
    $$
    \begin{aligned}
    \lim \left(\frac{1+2 a_{n}}{1+3 a_{n}-4 a_{n}^{2}}ight)
    &= \frac{\lim\left(1 + 2a_night)}{\lim \left(1 + 3a_n - 4a_n^2ight)} \
    &= \frac{1 + 2\lim (a_n)}{1 + 3\lim a_n - 4 \lim a_n^2} \
    &= 1
    \end{aligned}
    $$
    Showing $a_n^2 	o 0$ is easy so I've omitted it
  \item Apply the ALT
    $$
    \begin{aligned}
    \lim \left(\frac{\left(a_{n}+2ight)^{2}-4}{a_{n}}ight)
    &= \lim \left(\frac{a_{n}^2 + 2a_n}{a_{n}}ight) \
    &= \lim(a_n + 2) = 2 + \lim a_n = 2
    \end{aligned}
    $$
  \item Multiply the top and bottom by $a_n$ then apply the ALT
    $$
    \begin{aligned}
    \lim \left(\frac{\frac{2}{a_{n}}+3}{\frac{1}{a_{n}}+5}ight)
    &= \lim \left(\frac{2 + 3a_n}{1 + 5a_n}ight) \
    &= \frac{2 + 3 \lim a_n}{1 + 5 \lim a_n} \
    &= 2
    \end{aligned}
    $$
   \end{enumerate}

Exercise 2.3.4

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Exercise 2.3.4

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