Introduction to Limits in Calculus
Explore the fundamental concept of limits in calculus, essential for understanding slopes, lengths of curves, and areas of regions. Learn about the importance and informal description of limits, interpretation of limit notations, and how to calculate limits using the approach to a single number.
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Session Three Limits and Continuity An Introduction to Limits Calculating Limits using the Limit Laws
An Introduction to Limits We could begin by saying that limits are important in calculus, but that would be a major understatement. Without limits, calculus would not exist. Every single notion of calculus is a limit in one sense or another. For example: What is the slope of a curve? It is the limit of slopes of secant lines. What is the length of a curve? It is the limit of the lengths of polygonal paths inscribed in the curve. What is the area of a region bounded by a curve? It is the limit of the sum of areas of approximating rectangles.
An Introduction to Limits James Stewart, Calculus: Early Transcendentals, 7th Edition, Brooks/ Cole 2012. Howard Anton, Irl C. Bivens and Stephen Davis, Calculus: Early Transcendentals, 9th Edition, John Wiley & Sons, Inc. 2010. Salas, Etgen, and Hille, Calculus: One and Several Variables, 10th Edition, John Wiley & Sons, Inc. 2007.
An Introduction to Limits The informal description of a limit is as follows: If ?(?) becomes arbitrarily close to a single number ? as ? approaches ? from either side, the limit of ?(?), as ? approaches ? ,is ? . The existence or nonexistence of ?(?) at ? = ? has no bearing on the existence of the limit of ?(?) as ? approaches ? .
An Introduction to Limits The notation for a limit is lim ? ?? ? = ? which is read as the limit of ?(?) as ? approaches ? is ? . ? ? ?? ?. Example: Guess the value of lim ? ? ? ? ?? ?is not Solution: Notice that the function lim ? ? defined when ? = ?, but that does not matter because the definition of lim consider values of ? that are close to ? but not equal to ? . ? ??(?) says that we
An Introduction to Limits Continue: The table below gives values of (correct to six decimal places) for values of that approach (but are not equal to ). On the basis of the values in the table, we make the guess that ? ? ?? ?=? lim ? ? ?
An Introduction to Limits Definition: We write lim ? ? ?(?) = ? and say the left- hand limit of ?(?) as ? approaches ? is equal to ? if we can make the values of ?(?) arbitrarily close to ? by taking ? to be sufficiently close to ? and ?less than ? . Similarly, if we require that ? be greater than ? , we get the right-hand limit of ?(?) as ? approaches ? is equal to ? , and we write lim ? ?+?(?) = ? .
An Introduction to Limits Thus the symbol ? ? means that we consider only ? < ? , and the symbol ? ?+means that we consider only ? > ? . The relationship between ordinary (two-sided) limits and one-sided limits can be stated as follows. Theorem: Let ? . We say lim equals to ? if and only if lim ? ? ?(?) = lim ? ??(?) exists and ? ?+?(?) = ?
An Introduction to Limits Limits That Fail to Exist: o If a function ?(?) approaches a different number from the right side of ? = ? than it approaches from the left side, then the limit of ?(?) as ? approaches ? does not exist. lim ? ? ?(?) lim o If ?(?) is not approaching a real number ? that is, if ?(?) increases or decreases without bound as ? approaches ? , you can conclude that the limit does not exist. ? ?+?(?) lim ? ??(?) = o The limit of ?(?) as ? approaches ? does not exist if ?(?) oscillates between two fixed values as ? approaches ? .
An Introduction to Limits Example: The graph of a function ? is shown in the figure. Use it to state the value (if it exists) of lim ? ??(?) . Solution: From the graph we see that the values of ? ? approach ? as ? approaches ? from the left, but they approach ? as ? approaches ? from the right. Therefore lim ? ? ?(?) = ? and lim limits are different, we conclude that lim ? ??(?) does not exist. ? ?+?(?) = ? . Since the left and right
Calculating Limits using the Limit Laws In this section we use the following properties of limits, called the Limit Laws, to calculate limits. Theorem: Suppose that ? is a constant and the limits lim ? ??(?) and lim ? ??(?) exist. Then ?. lim ? ?[? ? + ? ? ] = lim ?. lim ? ?[? ? ? ? ] = lim ?. lim ? ?[?? ? ] = ? lim ?. lim ? ?[? ? ? ? ] = lim ? ? ? ? = ? ??(?) + lim ? ??(?) lim ? ??(?). ? ??(?) lim ? ?? ? lim ? ??(?). ? ?? ? . ? ??(?). lim ? ?? ??? lim ?. lim ? ? ? ??(?) ?.
Calculating Limits using the ? ?[? ? ]?= [lim 7. lim ? ?? = ? 8. lim ? ?? = ? 9. lim ? ??(?)]?? ??? ? ?? ? ???????? ??????? 6. lim ? ???= ??? ??? ? ?? ? ???????? ??????? 10. lim ? ? ??? ?? ? ?? ????,?? ?????? ? ?? ? ?? = ?? ? ??? ? ?? ? ???????? ???????, ??(?) = ?lim ? ??(?)? ??? ? ?? ? ???????? ??????? 11. lim ? ? ,??? ?? ? ?? ????,?? ?????? ? ?? lim ? ??(?) > 0
Example: Given that lim = ? , find lim Solution: lim ? ?.?[? ? ? ? ] = lim = lim ? ?.??(?) ? lim ? ?.??(?) = ? and lim ? ?.?[? ? ? ? ] . ? ?.??(?) ? ?.??(?) lim ? ?.?[?(?)] = ? ? = ? ? ?.?[?? ? ] Direct Substitution Property: If is a polynomial or a rational function and is in the domain of ? , then lim ? ??(?) = ?(?) . For example, lim = ( ?)??= ? ? ?(?? ???+ ?)??= [?? ? ??+ ?]??
The next examples show various ways algebraic manipulations can be used to evaluate lim situations where ?(?) is undefined. This usually happens when ?(?) is a fraction with denominator equal to ? at ? = ? . ?? ? ? ? Solution: We factor the numerator as a difference of squares:(?+?)(? ?) ? ? . The numerator and denominator have a common factor of ? ? . Therefore, ?? ? ? ?= lim = ? ? ??(?) in Example: Evaluate lim ? ? ? ?:(? + ?)(? ?) ? ? lim ? ? = lim ? ?? + ?
???? ?? Example: Find lim ? ? Solution: Here the preliminary algebra consists of rationalizing the numerator: ?? ? ? ?? ? ? ??+ ? ? ?? ? ? ?? ??+ ? + ? lim ? ? = lim . ??+ ? + ? ?? = lim ? ? = lim ? ? ??( ??+ ? + ?) ??( ??+ ? + ?) =? ? ? = lim ? ? ??+ ? + ?
?? ?? ? Example: Find lim ? ? Solution: Here the preliminary algebra consists of rationalizing both the numerator and the ???+??+? ???+??+?which ? ? ?? ? denominator by multiplying takes long time in calculations. The Substitution method is much better in this example. The idea is: write the problem using other variable ? so that the problem will transform to nice form that can easily solve. So, let ? = ??where ? = ??? ?,? .
?? = ? . Continue: Also, as ? ? we have ? Hence, ? ? ?? ?= lim ?? ? ?? ? ?? ? lim ? ? = lim ? ? ??? ? ? 1 (? ?)(??+ ? + ?) (? ?)(? + ?) ??+ ? + ? ? + ? = lim ? ? =? = lim ? ? ?
