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

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

1. 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.
2. 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 $$[4 - \delta, 4 + \delta]$$.
3. Given $$f(x) = 5 + \frac{3}{2 + x}$$. If $$y = L$$ is the equation of the horizontal asymptote of $$f$$,
1. find $$L$$
2. Solution As $$x \to \infty \text{, } 2 + x \to \infty \text{, so } \frac{3}{2 + x} \to 0$$. Therefore, $$f(x) \to 5$$.
3. find the value $$M$$ so that when $$x < M$$, $$|f(x) - L| < 0.05$$.
4. 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.
4. Given $$\displaystyle f(x) = \frac{1}{(2x - 8)^2}$$, find the value of $$\delta$$ so that $$f(x) > 150$$ when $$4 - \delta < x < 4 + \delta$$.
5. 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.