\begin{deloppgaver}
  \delo
    \begin{align*}
      a_n &= \frac{n}{n+1} \\[1ex]
      a_1 &= \frac{1}{1+1} = \frac{1}{2} \\[1ex]
      a_2 &= \frac{2}{2+1} = \frac{2}{3} \\[1ex]
      a_3 &= \frac{3}{3+1} = \frac{3}{4} \\[1ex]
      a_4 &= \frac{4}{4+1} = \frac{4}{5} \\[1ex]
      a_{100} &= \frac{100}{100+1} = \frac{100}{101} \\[1ex]
    \end{align*}

    Jeg gjetter at \[\lim_{n \to \infty} a_n = 1\]

  \delo
    \begin{align*}
      a_n &= \frac{n^2}{n+1} \\[1ex]
      a_1 &= \frac{1^2}{1+1} = \frac{1}{2} \\[1ex]
      a_2 &= \frac{2^2}{2+1} = \frac{4}{3} \\[1ex]
      a_3 &= \frac{3^2}{3+1} = \frac{9}{4} \\[1ex]
      a_4 &= \frac{4^2}{4+1} = \frac{16}{5} \\[1ex]
      a_{100} &= \frac{100^2}{100+1} = \frac{10000}{101} \\[1ex]
    \end{align*}

    Jeg gjetter at \[\lim_{n \to \infty} a_n = \infty\]

  \delo
    \begin{align*}
      a_n &= \frac{n}{n^2+1} \\[1ex]
      a_1 &= \frac{1}{1^2+1} = \frac{1}{2} \\[1ex]
      a_2 &= \frac{2}{2^2+1} = \frac{2}{5} \\[1ex]
      a_3 &= \frac{3}{3^2+1} = \frac{3}{10} \\[1ex]
      a_4 &= \frac{4}{4^2+1} = \frac{4}{17} \\[1ex]
      a_{100} &= \frac{100}{100^2+1} = \frac{100}{10001} \\[1ex]
    \end{align*}

    Jeg gjetter at \[\lim_{n \to \infty} a_n = 0\]
\end{deloppgaver}