Answers to exercises

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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