Limits and Continuity

\(y = -\dfrac{1}{3}x + \dfrac{2}{3}\)

\(y = \dfrac{2}{3}x + \dfrac{1}{3}\)

\(y = \dfrac{4}{3}x - \dfrac{3}{3}\)

\(y = \dfrac{1}{3}x \)

\(\displaystyle y = \frac{2}{3}x - \frac{1}{3}\)

If \(n=2\), then \(\lim\limits_{t \to 0^+} q(t) = 1/2\), meaning the point \(Q\) will approach the point \((0,1/2)\). On the other hand, if \(n >2\), then \(\lim\limits_{t \to 0^+} q(t) = + \infty\), meaning the point \(Q\) will rise indefinitely.

If \(n=2\), then \(\lim\limits_{t \to 0^+} q(t) = 0\), meaning the point \(Q\) will approach the point \((0,0)\). On the other hand, if \(n >2\), then \(\lim\limits_{t \to 0^+} q(t) = + \infty\), meaning the point \(Q\) will rise indefinitely.

If \(n=2\), then \(\lim\limits_{t \to 0^+} q(t) = 1\), meaning the point \(Q\) will approach the point \((0,1)\). On the other hand, if \(n >2\), then \(\lim\limits_{t \to 0^+} q(t) = + \infty\), meaning the point \(Q\) will rise indefinitely.

For each \(n>1\), we have \(\lim\limits_{t \to 0^+} q(t) = 1/n\), meaning the point \(Q\) will approach the point \((0,1/n)\).

For any \(n>1\), we have \(\lim\limits_{t \to 0^+} q(t) = + \infty\), meaning the point \(Q\) will rise indefinitely.

We have that \(\lim\limits_{r \to 0^+} p(r) = 2\), that is, point \(P\) will approach point \((2,0)\) as \(r\) approaches zero.

We have that \(\lim\limits_{r \to 0^+} p(r) = 0\), that is, point \(P\) will approach point \((0,0)\) as \(r\) approaches zero.

We have that \(\lim\limits_{r \to 0^+} p(r) = 4\), that is, point \(P\) will approach point \((4,0)\) as \(r\) approaches zero.

We have that \(\lim\limits_{r \to 0^+} p(r) = + \infty\), that is, point \(P\) will move indefinitely to the right as \(r\) approaches zero.

We have that \(\lim\limits_{r \to 0^+} p(r) = 8\), that is, point \(P\) will approach point \((8,0)\) as \(r\) approaches zero.

The only vertical asymptote of \(f\) is the line \(x = 0\), while \(f\) has no horizontal asymptotes.

The only vertical asymptote of \(f\) is the line \(x = 0\), and the only horizontal asymptote is the line \(y = 0\).

We have that \(f\) has no vertical asymptotes and has a single horizontal asymptote, which is the line \(y=0\).

We have that \(f\) has no vertical asymptotes and has a single horizontal asymptote, which is the line \(y=1\).

We have that \(f\) has neither horizontal nor vertical asymptotes.

\(a = - 16 \ \ \ \) and \( \ \ \ b = - \dfrac{1}{8}.\)

\(a = - 16 \ \ \ \) and \( \ \ \ b = \dfrac{1}{8}.\)

\(a = 8 \ \ \ \) and \( \ \ \ b = - \dfrac{1}{8}.\)

\(a = - 8 \ \ \ \) and \( \ \ \ b = - \dfrac{1}{8}.\)

\(a = 16 \ \ \ \) and \( \ \ \ b = - \dfrac{1}{8}.\)

Suppose \(n\) is odd. If \(a_n>0\), we have \(\lim\limits_{x \to - \infty} p(x) = -\infty\) and \(\lim\limits_{x \to + \infty} p(x) = + \infty.\) However, if \(a_n<0\), we have \(\lim\limits_{x \to - \infty} p(x) = + \infty\) and \(\lim\limits_{x \to + \infty} p(x) = +\infty\). Therefore, it is not possible to affirm that every polynomial of odd degree with real coefficients has a real root.

Suppose \(n\) is even. If \(a_n>0\), we have \(\lim\limits_{x \to - \infty} p(x) = -\infty\) and \(\lim\limits_{x \to + \infty} p(x) = +\infty\). However, if \(a_n<0\), we have \(\lim\limits_{x \to - \infty} p(x) = + \infty\) and \(\lim\limits_{x \to + \infty} p(x) = -\infty.\) In any case, there exist \(a < b\) such that \(p(a)\) and \(p(b)\) have opposite signs. Since \(p(x)\) is continuous on \(\mathbb{R}\), and therefore on \([a,b]\), it follows from the Intermediate Value Theorem that there exists at least one point \(x_0 \in (a,b)\) such that \(p(x_0) = 0\). In summary, every polynomial of even degree with real coefficients has at least one real root.

Suppose \(n\) is odd. If \(a_n>0\), we have \(\lim\limits_{x \to - \infty} p(x) = \infty\) and \(\lim\limits_{x \to \infty} p(x) = -\infty\). However, if \(a_n<0\), we have \(\lim\limits_{x \to - \infty} p(x) = -\infty\) and \(\lim\limits_{x \to +\infty} p(x) = +\infty\). In any case, there exist \(a < b\) such that \(p(a)\) and \(p(b)\) have opposite signs. Since \(p(x)\) is continuous on \(\mathbb{R}\), and therefore on \([a,b]\), it follows from the Intermediate Value Theorem that there exists at least one point \(x_0 \in (a,b)\) such that \(p(x_0) = 0\). In summary, every polynomial of odd degree with real coefficients has at least one real root.

Suppose \(n\) is odd. If \(a_n>0\), we have \(\lim\limits_{x \to - \infty} p(x) = -\infty\) and \(\lim\limits_{x \to + \infty} p(x) = + \infty\). However, if \(a_n<0\), we have \(\lim\limits_{x \to - \infty} p(x) = +\infty\) and \(\lim\limits_{x \to \infty} p(x) = -\infty\). In any case, there exist \(a < b\) such that \(p(a)\) and \(p(b)\) have opposite signs. Since \(p(x)\) is continuous on \(\mathbb{R}\), and therefore on \([a,b]\), it follows from the Intermediate Value Theorem that there exists at least one point \(x_0 \in (a,b)\) such that \(p(x_0) = 0\). In summary, every polynomial of odd degree with real coefficients has at least one real root.

Suppose \(n\) is even. If \(a_n>0\), we have \(\lim\limits_{x \to - \infty} p(x) = + \infty\) and \(\lim\limits_{x \to + \infty} p(x) = -\infty.\) However, if \(a_n<0\), we have \(\lim\limits_{x \to - \infty} p(x) = -\infty\) and \(\lim\limits_{x \to + \infty} p(x) = + \infty\). In any case, there exist \(a < b\) such that \(p(a)\) and \(p(b)\) have opposite signs. Since \(p(x)\) is continuous on \(\mathbb{R}\), and therefore on \([a,b]\), it follows from the Intermediate Value Theorem that there exists at least one point \(x_0 \in (a,b)\) such that \(p(x_0) = 0\). In summary, every polynomial of even degree with real coefficients has at least one root.