#CALCULUS LIMITS HOW TO#

lim x → 2 x + lim x → 2 1 ( lim x → 2 x ) 3 + lim x → 2 4 Apply the power law.lim x → 2 x + lim x → 2 1 lim x → 2 x 3 + lim x → 2 4 Apply the sum law and constant multiple law.Lim x → 2 2 x 2 − 3 x + 1 x 3 + 4 = lim x → 2 ( 2 x 2 − 3 x + 1 ) lim x → 2 ( x 3 + 4 ) Apply the quotient law, making sure that. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied. To find this limit, we need to apply the limit laws several times. We now practice applying these limit laws to evaluate a limit. Root law for limits: lim x → a f ( x ) n = lim x → a f ( x ) n = L n lim x → a f ( x ) n = lim x → a f ( x ) n = L n for all L if n is odd and for L ≥ 0 L ≥ 0 if n is even and f ( x ) ≥ 0 f ( x ) ≥ 0. Power law for limits: lim x → a ( f ( x ) ) n = ( lim x → a f ( x ) ) n = L n lim x → a ( f ( x ) ) n = ( lim x → a f ( x ) ) n = L n for every positive integer n. Quotient law for limits: lim x → a f ( x ) g ( x ) = lim x → a f ( x ) lim x → a g ( x ) = L M lim x → a f ( x ) g ( x ) = lim x → a f ( x ) lim x → a g ( x ) = L M for M ≠ 0 M ≠ 0 Product law for limits: lim x → a ( f ( x ) lim x → a f ( x ) = c L lim x → a c f ( x ) = c.Sum law for limits: lim x → a ( f ( x ) + g ( x ) ) = lim x → a f ( x ) + lim x → a g ( x ) = L + M lim x → a ( f ( x ) + g ( x ) ) = lim x → a f ( x ) + lim x → a g ( x ) = L + Mĭifference law for limits: lim x → a ( f ( x ) − g ( x ) ) = lim x → a f ( x ) − lim x → a g ( x ) = L − M lim x → a ( f ( x ) − g ( x ) ) = lim x → a f ( x ) − lim x → a g ( x ) = L − MĬonstant multiple law for limits: lim x → a c f ( x ) = c

Then, each of the following statements holds: Assume that L and M are real numbers such that lim x → a f ( x ) = L lim x → a f ( x ) = L and lim x → a g ( x ) = M. Let f ( x ) f ( x ) and g ( x ) g ( x ) be defined for all x ≠ a x ≠ a over some open interval containing a. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions. The first two limit laws were stated in Two Important Limits and we repeat them here. These two results, together with the limit laws, serve as a foundation for calculating many limits. We begin by restating two useful limit results from the previous section. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes.

#CALCULUS LIMITS HOW TO#

In this section, we establish laws for calculating limits and learn how to apply these laws. In the previous section, we evaluated limits by looking at graphs or by constructing a table of values.

2.3.6 Evaluate the limit of a function by using the squeeze theorem.

2.3.5 Evaluate the limit of a function by factoring or by using conjugates.

2.3.4 Use the limit laws to evaluate the limit of a polynomial or rational function.

2.3.3 Evaluate the limit of a function by factoring.

2.3.2 Use the limit laws to evaluate the limit of a function.

Calculus limits