UNIT:1 LIMITS &CONTINUITY (FOR BCA 1st SEMESTER)

Limits We write lim x→a f(x) = L or f(x) → L as x → a to mean that f(x) approaches the number L as x approaches (but is not equal to) a from both sides. A more precise way of phrasing the definition is that we can make f(x) be as close to L as we like by making x be sufficiently close to a. We write lim x→a+ f(x) = L or f(x) → L as x → a+ and lim x→a- f(x) = L or f(x) → L as x → a- to mean that f(x) → L as x approaches a from the right or left, respectively. For limx → a f(x) to exist, the left and right limits must both exist and must be equal. We write lim x→+∞ f(x) = L and lim x→-∞ f(x) = L to mean that f(x) → L as x gets arbitrarily large or becomes a negative number with arbitrarily large magnitude, respectively.	Examples 1. As x → 3, the quantity 3x2-4x+2 approaches 17, and hence lim x→3 (3x2-4x+2) = 27 Notice this is just the value of the function at x = 3 (see "Algebraic Approach" below). 2. Notice that the function x2 - 9 x - 3 is not defined at x = 3. However, for other values of x, it simplifies to x2 - 9 x - 3 = (x - 3)(x + 3) x - 3 = x + 3, and, as x → 3, this quantity approaches 6. Therefore x2 - 9 x - 3 → 6   as  x → 3, or lim x→3 x2 - 9 x - 3 = 6 There are many more examples in the on-line tutorials on limits. Top of Page
Estimating Limits Numerically To analyze a limit of the form lim x→a f(x)     or     lim x→±∞ f(x)   numerically: Make a table of values for f(x) using values of x that approach a closely from either side. If the limit exists, then the values of f(x) will approach the limit as x approaches a from both sides. The more accurately you wish to evaluate this limit, the closer to a you will need to choose the values of x. For a limit as x → +∞, use positive values of x getting larger and larger. For a limit as x → -∞, use negative values of x getting larger and larger in magnitude.	Examples 1. To estimate lim x→3 x2 - 9 x - 3 , constuct a table with values of x close to 3 on either side: x approaching 3 from the left →    x approaching 3 from the right ← x 2.9 2.99 2.999 2.9999 f(x) = x2 - 9 x - 3 5.9 5.99 5.999 5.9999 3 3.0001 3.001 3.01 3.1 6.0001 6.001 6.01 6.1 Since the values of f(x) appear to be approaching 6 as x approaches 3 from either side, we estimate that the limit is 6. 2. To estimate lim x→ +∞ x2 - x + 1 2x2 - 3 , constuct a table with values of x approaching +∞: x approaching +∞   → x 10 100 1000 10,000 f(x) = x2 - x + 1 2x2 - 3 0.461929 0.495124 0.499501 0.49995 +∞ Since the values of f(x) appear to be approaching 0.5 as x approaches 3 from either side, we estimate that the limit is 0.5. Top of Page
Estimating Limits Geometrically To analyze a limit of the form lim x→a f(x)     or     lim x→±∞ f(x)   geometrically: Draw the graph of f(x) either by hand or using technology, such as a graphing calculator. If you want to calculate a limit as x → a for a real number a, position your pencil point (or the graphing calculator "trace" cursor) on a point of the graph to the left of x = a. Move the point along the graph toward x = a from the left and read the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit lim x→a- f(x) Repeat the above two steps, but this time starting from a point on the graph to the right of x = a, and approach x = a along the graph from the right. The value the y-coordinate approaches (if any) is then lim x→a+ f(x) If the left and right limits both exist and have the same value L, then lim x→a+ f(x) exists and equals L. If you want to calculate a limit as x → +∞, position your pencil point (or the graphing calculator "trace" cursor) on a point near the right edge of the graph, and move the pencil along the graph to the right, estimating the y-coordinate as you go. The value the y-coordinate approaches (if any) is then the limit lim x→+∞+ f(x) For x → -∞, start toward the left edge, and move your pencil toward the left.	Example The following picture shows the graph of f(x). First, we calculate the limit of f(x) as x → 0 from the left: lim x→0- f(x)   Thus,   lim x→0- f(x) = -1 Next, we calculate the limit of f(x) as x → 0 from the right: lim x→0+ f(x)   Thus,   lim x→0+ f(x) = 3 Since the left- and right limits disgaree, we conclude that lim x→0 f(x) does not exist. Top of Page
Computing Limits Algebraically: Limits as x → a To compute a limit of the form lim x→a f(x)   algebraically: Check to see whether f is a closed form function. These are functions specified by a single formula involving constants, powers of x, radicals, exponentials and logarithms, combined using arithmetic operations and composition of functions. If a is in the domain of f, then limx → a f(x) = f(a). If a is not in the domain of f, but f(x) can be reduced by simplification to a function with a in its domain, then (a) applies to the reduced form of the function. If a is not in the domain of f, and you cannot simplify the function as in (b), then simplify as much as possible, and evaluate the limit by the numerical approach, or, if you know its graph, by the graphical approach. If f is not closed-form, and a is a point at which the definition of f changes, compute the left limit and right limit seperately, and check whether they agree. Limits as x → ±∞ To compute a limit of the form lim x→±∞ f(x)   algebraically: If x is approaching ±∞, then check to see whether f(x) is a ratio of polynomials. If it is, then you can ignore all but the highest powers of x in the numerator and denominator. This simpler function will have the same limit as f.	Examples 1. Consider the limit lim x→1 x2 - 9 x - 3 . Notice that the function f(x) = x2 - 9 x - 3 is closed form, and a = 1 is in its domain. Therefore, the limit is obained by substituting x = 1 (point (a) opposite): lim x→1 x2 - 9 x - 3 = 1 - 9 1 - 3 = 4 2. Now consider lim x→3 x3 - 9 x - 3 . This time, a = 3, which is not in the domain of f, and so we need to first simplify f(x) to reduce it to a function that does have 3 in its domain: lim x→3 x2 - 9 x - 3 = lim x→3 (x - 3)(x + 3) x - 3 = lim x→3 x + 3. Now, x = 3 is in the domain of f, and so we find the limit by putting x = 3: lim x→3 x2 - 9 x - 3 = lim x→3 x + 3 = 3 + 3 = 6. 3. Consider lim x→+∞ x3 + x2 - 9 2x3 - x - 3 . Here f(x) is a ratio of polynomials, so we ignore everything except the highest power of x in the numerator and denominator: lim x→+∞ x3 + x2 - 9 2x3 - x - 3 = lim x→+∞ x3 + x2 - 9 2x3 - x - 3 = lim x→+∞ x3 2x3 = lim x→+∞ 1 2 = 1 2 (Cancel the x2) Press here for the on-line tutorial on limits. Top of Page
Continuous Functions A function f is continuous at a if limx → a f(x) exists, and is equal to f(a). The function f is said to be continuous on its domain if it is continuous at each point in its domain. The algebraic approach to limits above is based on the fact that any closed form function is continuous on its domain.	Examples The function f(x) = 3x2-4x+2 is a closed form function, and hence continuous at every point in its domain (all real numbers). The function g(x) = 4x2+1 x - 3 is also a closed form function, and hence continuous on its domain (all real numbers excluding 3). On the other hand, the function h(x) = -1 if -4 ≤ x < -1 x if -1 ≤ x ≤ 1 x2-1 if  1 < x ≤ 2  is not a closed-form function, and is in fact discontinuous at x = 1.