Limits and Continuity | q002.html

(cod: P-44-14-15) Let \(P\) be a point in the first quadrant on the curve \(y = x^n\), where n is a natural number greater than 1. Consider the segment \(\overline{OP}\) that connects the point \(P\) to the origin \(O\), and let \(M\) denote the midpoint of this segment. Consider the line \(r\) that is orthogonal to the segment \(\overline{OP}\) passing through \(M\), and let \(Q\) be the point given by the intersection of \(r\) with the \(y\)-axis. The figure below illustrates the situation in the case \(n = 2\):

We want to determine what will happen to the point \(Q\) if we bring the point \(P\) closer to the origin, maintaining the described structure. Writing \(P = (t,t^n)\) and \(Q = (0,q(t))\), select the correct alternative:

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.