[proofplan] We separate the constant polynomial case from the positive-degree case. When $d=0$, the quotient is identically $1$. When $d \ge 1$, we divide the polynomial by its leading term and obtain $1$ plus finitely many lower-order terms of the form $c_i x^{i-d}$, where $i-d<0$. We then prove directly from the definition of the limit as $|x| \to \infty$ that the finite lower-order sum tends to $0$. [/proofplan] [step:Handle the constant polynomial case] Suppose first that $d=0$. Then $p(x)=a_0$ for every $x \in \mathbb{R}$, and $a_d=a_0 \ne 0$. Since $x^0=1$ for every $x \in \mathbb{R}$, we have \begin{align*} \frac{p(x)}{a_d x^d}=\frac{a_0}{a_0}=1 \end{align*} for every $x \in \mathbb{R}$. Hence \begin{align*} \lim_{|x| \to \infty} \frac{p(x)}{a_d x^d}=1. \end{align*} [/step] [step:Rewrite the normalized polynomial as one plus lower-order terms] Assume now that $d \ge 1$. For every $x \in \mathbb{R}$ with $|x| \ge 1$, we have $x \ne 0$, so $a_d x^d \ne 0$ because $a_d \ne 0$. Therefore the quotient is defined for all $x$ with $|x| \ge 1$. For each integer $i \in \{0,\dots,d-1\}$, define the real constant \begin{align*} c_i=\frac{a_i}{a_d}. \end{align*} Then, for every $x \in \mathbb{R}$ with $|x| \ge 1$, \begin{align*} \frac{p(x)}{a_d x^d}=1+\sum_{i=0}^{d-1} c_i x^{i-d}. \end{align*} [guided] The purpose of normalizing by $a_d x^d$ is to isolate the leading term. Since $d \ge 1$, the denominator $a_d x^d$ is nonzero whenever $x \ne 0$; in particular it is nonzero for all $x$ satisfying $|x| \ge 1$. For each integer $i \in \{0,\dots,d-1\}$, define \begin{align*} c_i=\frac{a_i}{a_d}. \end{align*} This is well-defined because the leading coefficient satisfies $a_d \ne 0$. Dividing the expression \begin{align*} p(x)=a_d x^d+\sum_{i=0}^{d-1} a_i x^i \end{align*} by $a_d x^d$ gives \begin{align*} \frac{p(x)}{a_d x^d}=1+\sum_{i=0}^{d-1} \frac{a_i}{a_d} x^{i-d}. \end{align*} Using the constants $c_i$ just defined, this becomes \begin{align*} \frac{p(x)}{a_d x^d}=1+\sum_{i=0}^{d-1} c_i x^{i-d}. \end{align*} Every exponent $i-d$ in the remaining sum is negative, because $i \le d-1$. Thus each term is a lower-order correction to the leading term. [/guided] [/step] [step:Show the lower-order finite sum tends to zero] Define the function $r: \mathbb{R} \setminus \{0\} \to \mathbb{R}$ by \begin{align*} r(x)=\sum_{i=0}^{d-1} c_i x^{i-d}. \end{align*} We prove that $\lim_{|x| \to \infty} r(x)=0$. Let $\varepsilon>0$. For each $i \in \{0,\dots,d-1\}$, define \begin{align*} b_i=|c_i|. \end{align*} Define the nonnegative constant \begin{align*} B=\sum_{i=0}^{d-1} b_i. \end{align*} If $B=0$, then $r(x)=0$ for every $x \ne 0$, so the desired limit is immediate. Assume $B>0$. Define \begin{align*} R=\max\{1,B/\varepsilon\}. \end{align*} If $|x|>R$, then $|x|>1$. Since $i-d \le -1$ for every $i \in \{0,\dots,d-1\}$, we have \begin{align*} |x^{i-d}|=|x|^{i-d}\le |x|^{-1}. \end{align*} By the triangle inequality, \begin{align*} |r(x)|\le \sum_{i=0}^{d-1} |c_i| |x^{i-d}|. \end{align*} Using the preceding bound term by term gives \begin{align*} |r(x)|\le \sum_{i=0}^{d-1} b_i |x|^{-1}=B|x|^{-1}<\varepsilon. \end{align*} Thus, for every $\varepsilon>0$, there exists $R>0$ such that $|x|>R$ implies $|r(x)|<\varepsilon$. Therefore \begin{align*} \lim_{|x| \to \infty} r(x)=0. \end{align*} [/step] [step:Conclude that the normalized polynomial tends to one] For every $x \in \mathbb{R}$ with $|x| \ge 1$, \begin{align*} \frac{p(x)}{a_d x^d}=1+r(x). \end{align*} Since $\lim_{|x| \to \infty} r(x)=0$, it follows directly from the definition of the limit that \begin{align*} \lim_{|x| \to \infty} \frac{p(x)}{a_d x^d}=1. \end{align*} Combining this with the already handled case $d=0$ proves the theorem. [/step]