##### Exercise 1
$$\displaystyle \lim_{n\to \infty} \dfrac{5n^3+(-1)^n}{4n^3+2} = \lim_{n\to \infty} \dfrac{5+\dfrac{(-1)^n}{n^3}}{4+\dfrac{2}{n^3}} = \dfrac{5}{4}$$.
##### Exercise 2

In the geometric series $$\dfrac{3}{2} + \dfrac{9}{8} + \dfrac{27}{32} + \dotsb$$, the first term $$a$$ is $$\dfrac{3}{2}$$ and the common ratio $$r$$ is $$\dfrac{3}{4}$$ (which is less than 1 in magnitude). So the limiting sum is

$S_\infty = \dfrac{a}{1-r} = \dfrac{\dfrac{3}{2}}{1-\dfrac{3}{4}} = 6.$
##### Exercise 3
$$\displaystyle \lim_{x\to \infty} \dfrac{x^2-1}{3x^2+1} = \lim_{x\to \infty} \dfrac{1-\dfrac{1}{x^2}}{3+\dfrac{1}{x^2}} = \dfrac{1}{3}$$.

So the function $$f(x)= \dfrac{x^2-1}{3x^2+1}$$ has horizontal asymptote $$y = \dfrac{1}{3}$$.

##### Exercise 4

Define $$f(x)= \dfrac{x^2}{x^2-1}$$. Then

$f(x) \to -\infty\ \text{as}\ x\to 1^- \qquad \text{and} \qquad f(x) \to +\infty\ \text{as}\ x\to 1^+,$

and so $$\lim\limits_{x\to 1} f(x)$$ does not exist. Also,

$f(x) \to +\infty\ \text{as}\ x\to -1^- \qquad \text{and} \qquad f(x) \to -\infty\ \text{as}\ x\to -1^+,$

hence $$\lim\limits_{x\to -1} f(x)$$ does not exist. We can calculate

$\lim_{x\to \infty} f(x) = \lim_{x\to \infty} \dfrac{1}{1-\dfrac{1}{x^2}} = 1 \qquad\text{and}\qquad \lim\limits_{x\to -\infty} f(x) = 1.$
##### Exercise 5

Define $$f(x) = \dfrac{(x-5)(x+3)}{(2x-1)(x+3)}$$. Then:

1. $$f(x) \to \dfrac{1}{2}$$ as $$x\to \infty$$
2. $$f(x) \to 0$$ as $$x\to 5$$
3. $$f(x) \to \dfrac{8}{7}$$ as $$x\to -3$$
4. $$f(x) \to -\infty$$ as $$x\to \dfrac{1}{2}^+$$, and $$f(x) \to +\infty$$ as $$x\to \dfrac{1}{2}^-$$, so $$f(x)$$ has no limit as $$x\to \dfrac{1}{2}$$
5. $$f(x) \to 5$$ as $$x\to 0$$.
##### Exercise 6
1. We rationalise the numerator: \begin{align*} \lim_{x\to 1} \dfrac{\sqrt{x^2+15}-4}{x-1} &= \lim_{x\to 1} \Biggl( \dfrac{\sqrt{x^2+15}-4}{x-1} \times \dfrac{\sqrt{x^2+15}+4}{\sqrt{x^2+15}+4} \Biggr)\\ &= \lim_{x\to 1} \dfrac{x^2-1}{(x-1)\bigl(\sqrt{x^2+15}+4\bigr)} = \lim_{x\to 1} \dfrac{(x-1)(x+1)}{(x-1)\bigl(\sqrt{x^2+15}+4\bigr)}\\ &= \lim_{x\to 1} \dfrac{x+1}{\sqrt{x^2+15}+4} = \dfrac{1}{4}. \end{align*}
2. We get rid of the fractions in the numerator: \begin{align*} \lim_{x\to 4} \dfrac{\dfrac{1}{x}-\dfrac{1}{4}}{x-4} &= \lim_{x\to 4} \Biggl( \dfrac{\dfrac{1}{x}-\dfrac{1}{4}}{x-4} \times \dfrac{4x}{4x} \Biggr) = \lim_{x\to 4} \dfrac{4-x}{4x(x-4)}\\ &= \lim_{x\to 4} \dfrac{-(x-4)}{4x(x-4)} = \lim_{x\to 4} \Bigl( -\dfrac{1}{4x}\Bigr) = -\dfrac{1}{16}. \end{align*}
##### Exercise 7

Clearly $$f(0) = 4$$. We need to look at the limit at $$x = 0$$ from above. For $$x>0$$, we have $$f(x) = 4+x$$. So $$f(x) \to 4$$ as $$x\to 0^+$$. Since this limit is equal to $$f(0)$$, we conclude that $$f$$ is continuous everywhere.

##### Exercise 8

Let $$\varepsilon > 0$$. We want to find $$M$$ such that, if $$x > M$$, then $$|f(x) - 2| < \varepsilon$$. Note that

$|f(x) - 2| = \Bigl|\dfrac{2x^2+3}{x^2}-2\Bigr| = \dfrac{3}{x^2}.$

We want $$\dfrac{3}{x^2} < \varepsilon$$, which is equivalent to $$x^2 > \dfrac{3}{\varepsilon}$$. Hence, we take $$M = \sqrt{\dfrac{3}{\varepsilon}}$$. For all $$x > M$$, we now have $$|f(x) - 2| = \dfrac{3}{x^2} < \varepsilon$$. This tells us that $$f(x)$$ has a limit of 2 as $$x\to \infty$$.

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