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Chapter 1: Limits
Section 1.2: Precise Definition of a Limit
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Example 1.2.8
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Use Definition 1.2.1 to verify , the limit explored in Example 1.2.3.
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Solution
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Type the equation
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Context Panel: Assign Function
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Figure 1.2.8(a) is an animation in which is graphed in blue, and , in black.
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The slider in the animation toolbar controls the value of . As the slider is moved past the first frame, red and green horizontal lines delineate an -band around and red and green vertical lines delineate the band .
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The red and green horizontal lines are drawn at , respectively, and the red and green vertical lines are drawn at the corresponding -coordinates and .
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f:=x->x^2-3*x+3:
g:=x->(3+sqrt(4*x-3))/2:
p1:=plot(f,2..4,color=black):
p2:=plots:-animate(plot,[[3+epsilon,3-epsilon,3],x=2..4,color=[green,red,blue]],epsilon=0..1,frames=26,background=p1):
p3:=plots:-animate(plot,[[[g(3+epsilon),t*f(g(3+epsilon)),t=0..1],[g(3-epsilon),t*f(g(3-epsilon)),t=0..1]],color=[green,red],titlefont=[Times,14]],epsilon=0..1,frames=26):
plots:-display(p||(1..3));
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Figure 1.2.8(a) Animation illustrating Definition 1.2.1
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Write the equation
Press the Enter key.
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Context Panel: Solve≻Obtain Solutions for≻
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Select the solution that tends to 0 as . (The other solution tends to 3.) Hence,
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Write the equation
Press the Enter key.
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Context Panel: Solve≻Obtain Solutions for≻
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Select the solution that tends to 0 as . (The other solution tends to .) Hence,
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Figure 1.2.8(b) provides graphical evidence that, for small values of ε, .
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For an analytic determination that , show .
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Begin by simplifying the ratio , look at its graph, and finish via the calculations in Table 1.2.8(a).
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use plots, plottools in
module()
local dL,dR,p1,p2,p3,ve;
ve:=varepsilon;
dL:=(3-sqrt(9-4*ve))/2;
dR:=(sqrt(9+4*ve)-3)/2;
p1:=plot([dR,dL],ve=0..1/2,color=[black,red],tickmarks=[5,3]);
p2:=textplot({[.4,.18,typeset(delta[L])],[.4,.085,typeset(delta[R])]});
p3:=display(p1,p2,scaling=constrained);
print(p3);
end module:
end use:
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Figure 1.2.8(b)
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=
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The first step in Table 1.2.8(a) is the rationalization of the denominator; the second is making the resulting numerator smaller by changing to in the second factor.
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=
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Table 1.2.8(a) Show
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Consequently,
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6
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Table 1.2.8(b) Verification that ⇒
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Lines (1) and (2) in Table 1.2.8(b) are obtained with basic algebra. The result in (3) is obtained by applying the triangle inequality: . The coefficients of and in (3) are positive (graph them!), so their absolute values are dropped in (4). Also in (4), replacing with makes the expression in (3) larger because for , . Simple arithmetic applied to (4) gives (5), and hence (6).
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