Solutions

(a) We can use the algebraic limit theorem to convince ourselves that

$$1 + 2a_{n} \rightarrow 1$$

by applying rule (i) on $2a_{n}$ and (ii) on $1 + 2a_{n}$. For the terms of the denominator, we first apply rule (i) on $3a_{n}$ to obtain

$$3a_{n} \rightarrow 0.$$

Next, we apply (iii) on $4a_{n}^2$ to obtain

$$4a_{n}^2 \rightarrow 0.$$

Lastly, we apply (ii) on the whole sum $1 + 3a_{n} - 4a_{n}^2$ to obtain

$$1 + 3a_{n} - 4a_{n}^2 \rightarrow 0.$$

Thus, putting everything together, we get

$$\begin{aligned} \lim \frac{1 + 2a_{n}}{1 + 3a_{n} - 4a_{n}^2} &= \frac{\lim (1 + 2a_{n})}{\lim ( 1 + 3a_{n} - 4a_{n}^2 )} \\ &= \frac{1 + 2(0)}{1 + 3(0) -4(0^2)} = 1. \end{aligned}$$

(b) Though this sequence seems to converge to an undefined value because its denominator is $a_{n}$, we can apply the the difference of squares rule to the numerator to reveal a simplifaction. That is,

$$\lim \frac{( a_{n} + 2 )^2 - 4}{a_{n}} = \lim \frac{a_{n} (a_{n} + 4)}{a_{n}} = \lim a_{n} + 4$$

Thus, by applying rule (ii) to the simplified limit, we realize that

$$\lim \frac{( a_{n} + 2 )^2 - 4}{a_{n}} = 4.$$

(c) This again seems like a sequence that converges to an undefined value. However, if we multiply the sequence by $a_{n}/a_{n}$, we can obtain a form easier to work with. That is,

$$\lim \frac{\frac{2}{a_{n}} + 3}{\frac{1}{a_{n}} + 5} = \lim \frac{\frac{2}{a_{n}} + 3}{\frac{1}{a_{n}} + 5} \cdot \frac{a_{n}}{a_{n}} = \lim \frac{2 + 3a_{n}}{1 + 5a_{n}}.$$

Applying rule (i) and (ii), this sequence obviously converges to 2.