Use the Limit Laws to evaluate limx rightarrow 2(2x + 3). The solution will rese

Question

Use the Limit Laws to evaluate limx rightarrow 2(2x + 3). The solution will resemble examples 1a, b, c, of the text, where the limit is evaluated in a sequence of steps with each step being justified by an appropriate Limit Law. Don't just write down an answer for the limit which we are all well aware is 7. The point of the problem is to understand how using just the few Limit Laws, we can evaluate the limit, Assume lim f(x) = 5. Use the Limit Laws to evaluate lim x rightarrow 3(f(x) + 7) lim x rightarrow 3(7 f(x)) lim x rightarrow 3 7 / f(x) lim x rightarrow 3 x + 7 / f(x) - 3 Suppose f(x) = x - 2 / x + 3. By computing an appropriate limit, show f is continuous at x = 1. Compute appropriate limits to show f(x) = x - 3 / |x - 3| has a jump discontinuity at x = 3. Suppose f(x) = On each of the intervals (-infinity, 2), (2, 4), and (4, infinity), f is a polynomial function, and so, according the the reading in section 2. 4, f is continuous on each of these intervals. Determine, by computing appropriate limits, if f is continuous at x = 2 and x = 4. Is f either left or right continuous (or both) at x = 2? Is f either left or right continuous (or both) at x = 4? Draw the graph of f. (Bonus question: worth 10 points. Total points for assignment not to exceed 100. ) Give and example of functions y = f(x) and y = g(x) such that lim x rightarrow 1 f(x) = 0 but lim x rightarrow 1 f(x) / g(x) = 8. The important point of such an example is that even if the numerator of a quotient has a limit of 0, you are not justified in concluding the limit of the quotient is 0.

Explanation / Answer

a) lim x->3 f(x) + 7 = lim x->3 f(x) + lim x->3 7 = 5 + 7 = 12 [lim x->k a + b = lim x->k a + lim x->k b] b) lim x->3 7f(x) = 7 lim x->3 f(x) = 7 * 5 =35 [ lim x->k pf(x) = p lim x->k p(x) ] c) lim x->3 7/f(x) = 7 (lim x->3 1)/(lim x->3 f(x)) = 7 * 1/5 = 7/5 d) lim x->3 (x+7)/(f(x) - 3) = (lim x->3 (x+7))/(lim x->3 f(x) - lim x->3 3) = 10/(5 - 3) = 10/2 = 5

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