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Graph f for m = 1, and find lim_x rightarrow 2^- f(x) and lim_x rightarrow 2^+ f

ID: 2874751 • Letter: G

Question

Graph f for m = 1, and find lim_x rightarrow 2^- f(x) and lim_x rightarrow 2^+ f(x) Find m so that lim_x rightarrow 2^- f(x) = lim_x rightarrow 2^+ f(x) and graph f for this value of m. Write a brief verbal description of each graph. He does the graph in part (C) differ from the graphs parts (A) and (B)? Find each limit in Problems 71-74, where a is a real constant. lim_h rightarrow 0 (a + h)^2 - a^2/h lim_h rightarrow 0 [3(a + h) - 2] - (3a - 2)/h lim_h rightarrow 0 squareroot a + h - squareroot a/h, a > 0 lim_h rightarrow 0 1/a + h - 1/a/h, a notequalto 0

Explanation / Answer

73)

limh->0[((a+h) -a)/h]

rationalise the numerater => multiply and divide by ((a+h) +a)

limh->0[((a+h) -a)/h]*[((a+h) +a)/((a+h) +a)]

limh->0[((a+h) -a)((a+h) +a)]/[h((a+h) +a)]

since (a-b)(a+b)=a2-b2

limh->0[((a+h)2 -a2)]/[h((a+h) +a)]

limh->0[((a+h) -a)]/[h((a+h) +a)]

limh->0[h]/[h((a+h) +a)]

limh->0 1/((a+h) +a)

=1/((a+0) +a)

=1/(a +a)

=1/(2a)

therefore limh->0[((a+h) -a)/h] =1/(2a)