# BC Calculus: Extra Limits Concept Practice. §2.2 and §2.3

- Class: 5H: BC Calculus
- Author: Peter Atlas
- Text:
__Calculus__ Finney, Demana, Waits, Kennedy

- Find a symmetric interval about \(x = 4\) so that so long as \(x\) is within that interval, \(f(x) = 0.2x^2 - \sqrt{x} + 1\) is within 0.1 units of its limiting value.
## Solution

Graph in calculator \(y_1 = 0.2x^2 - \sqrt{x} + 1\). Pick a window. I chose \(x: [3, 5]\) and guessed \(y: [0, 5]\). Press (trace) then 4 (enter) to find the value of \(f\) at \(x = 4\). It should tell you \(y = 2.2\). This is \(L\). Now graph \(y_2 = 2.2 - 0.1\), \(y_3 =2.2 + 0.1\), and change the window to \(x: [3, 5]\) by \(y: [2, 2.5]\). Find the intersection of \(y_2\) and \(y_1\) to get \(x = 3.925027\). Store this in \(A\). Now find the intersection of \(y_3\) and \(y_1\) to get \(x = 4.073218\). Store this in \(B\). So far, we know that when \(x\) is within \([A, B]\), the curve will be within 0.1 of its limit. But we need a *symmetric* interval. So I'll consider the length of the two radii: \(4 - A = 0.074972\) on the left, and \(B - 4 = 0.073218\) on the right. I'll let \(\delta\) = the minimum of these two numbers. So \(\delta = 0.073218\), and the symmetric interval is \([4 - \delta, 4 + \delta]\).
- Given \(f(x) = 5 + \frac{3}{2 + x}\). If \(y = L\) is the equation of the horizontal asymptote of \(f\),
- find \(L\)
## Solution

As \(x \to \infty \text{, } 2 + x \to \infty \text{, so } \frac{3}{2 + x} \to 0 \). Therefore, \(f(x) \to 5\).
- find the value \(M\) so that when \(x < M\), \(|f(x) - L| < 0.05\).
## Solution

To begin with, recognize that the question is asking about the behavior as \(x \to -\infty\). On the calculator, graph \(y_1 = 5 + \frac{3}{2 + x}\), \(y_2 = 5 - 0.05\), and \(y_3 = 5 + 0.05\) on the window \(x: [-100, 0]\text{ by }y: [4.9, 5.1]\). The intersection we want is with \(y_1\) and \(y_2\). Use the intersect function to get \(x = -62\). This is the \(M\) we were looking for.

- Given \( \displaystyle f(x) = \frac{1}{(2x - 8)^2}\), find the value of \(\delta\) so that \(f(x) > 150\) when \(4 - \delta < x < 4 + \delta\).
## Solution

Recognize that this question is asking about the limit on a vertical asymptote. Graph \( \displaystyle y_1 = \frac{1}{(2x - 8)^2}\text{ and }y_2 = 150\) on the window \(x: [3.9, 4.1] \text{ by }y: [0, 200]\). Find the intersections are \(x = \{3.9591752, 4.0408248 \} \), and store those in \(A\) and \(B\). \(x - A = 0.0408.\) \(B - x = 0.0408.\) Let \( \delta = 0.040.\) The answer is 0.040.