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Based on experience with y = 1/ x, this function has the same shape, it is just displaced (or translated) 1 unit to the right In y = 1/ x, x = 0 is the asymptote line, but in y = 1 / (x - I ) , x = 1 is the asymptote line Follow the logic of the limit calculations and verifL the graph as shown in Fig 2-3 Graph y = 1/ x2 using limit concepts and notation
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Solution: Think limits and write the symbolic statements
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Asx++a,y+O+
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ASX+0-, y + + a AS x + -00, y + O + Now draw in the curve (See Fig 2-4)
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x+2 x-5
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Fig 2-4
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Discuss the function y = ___ in the vicinity of x = 5
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Solution: The numerator of the function presents no problem Even at x = - 2 , the function is 0 / -7 = 0, perfectly understandable Based on past nvnn1;nnPm t m - C in tuiu UuiiviimiaLwi n r n A w - m c a h m Ammn-imotnr u ~ p ~ i i u i i ~hu , uiu ui pivuuuua o vertical asymptote at x = 5 Place the asymptote line on the graph Now, using limit language, describe the behavior of the function in the vicinity of x = 5
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A s x + ~ + , ~ + + ~ ox + ~ - , y + - a As
There is an additional complication as x becomes large, either positive or negative For large x the Fig 2-5 function becomes large number over large number If, however, the fraction is multiplied by 1/ x over 1/ x the limit can be calculated easily:
This limit produces a horizontal asymptote When x is greater than 5 (refer to the original function statement), the fraction is positive so this horizontal asymptote is approached fiom the positive side When x is less than 5, but greater than -2, the h c t i o n is negative At x = 0, y = -2 / 5 For values of x less than (to the left of) -2, the h c t i o n is
LIMITS AND CONTINUITY
always positive and for larger and larger negative x, the function approaches the limit 1 from the negative side Go through the logic and verie the graph of Fig 2-5
As the powers of the polynomials increase, the fimctions become harder to graph In chapter 4 more complicated polynomials will be graphed with the aid of calculus
Find the limit of the function y =
3x2+2x+1 xz+x+l
as x goes to infinity
Solution: Attempting to evaluate the b c t i o n for large x produces the result large number over large number Taking the limit with a little inventive algebra (multiplying the fiaction by I / x2 over 1 / x2) produces
x-+cr,
3x2 + 2 x + l x2+x+l
[G] E +
1/x2
3 + 2 / x -+ 1/x2
1 1/ x + 1/ x 2
Find the limit of
x4 + 3x2
as x goes to infinity
Solution: Again use a little inventive algebra Factor an x4 out of the numerator and an x5 out of the denominator: lim x 4 +3x2 = lim x4(1+ 3 / x 2 ) = lim[-l[ 1+3/x 2 ] = 0 1
x--)w
x5+2
~ + 0 0 ~ ~ ( 1 + 2 x+m ~x) 1+2/x5 /~
The fust fraction has limit zero and the second limit 1 The product is zero This problem illustrates a manipulative rule for limits lim A - B = lim A][ lim B ]
Continuous functions are defined mathematically, usually over specific intervals The requirements of a continuous function are: 1) it exists at every point in the defrned interval, and 2) the iimit exists at every point and is equal to the value of the function at that point Operationally, continuous functions are ones that can be drawn without lifting your pencil
CHAPTER2
Another often used sample of a discontinuous function is the integer function, y = [ x ] , where the [XI symbolism is read as "the largest integer contained in x" y = [XI For example, the largest integer contained in 2 is 2 The
Look back over problems 2-1 through 2-4 and note that the discontinuity occurs at the vertical asymptote Another example of a discontinuous function is one defined on certain intervals such as
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