Limits of algebraic functions *

Transcription

1 OpenStax-CNX module: m7542 Limits of algebraic functions * Sunil Kumar Singh This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 2.0 Algebraic expressions comprise of polynomials, surds and rational functions. For evaluation of its of algebraic functions, the main strategy is to work expression such that we get a form which is not indeterminate. Generally, it helps to know indeterminate form of expression as it is transformed in each step of evaluation process. The moment we get a determinate form, the it of the algebraic expression is obtained by plugging iting value of x in the expression. The approach to transform or change expression depends on whether independent variable approaches nite values or innity. The point of it determines the way we approach evaluation of it of a function. The treatment of its involving independent variable tending to innity is dierent and as such we need to distinguish these its from others. Thus, there are two categories of its being evaluated : : Limits of algebraic function when variable tends to nite value. 2: Limits of algebraic function when variable tends to innite Limits of algebraic function when variable tends to nite value In essence, we shall be using following three techniques to determine it of algebraic expressions when variable is approaching nite value not innity. These methods are : : Simplication or rationalization for radical functions) 2: Using standard it form 3: Canceling linear factors for rational function) We should be aware that if given function is in determinate form, then we need not process the expression and obtain it simply by plugging iting value of x in the expression. Some problems can be alternatively solved using either of above methods.. Simplication or rationalization for radical functions We simplify or rationalize if surds are involved) and change indeterminate form to determinate form. We need to check indeterminate forms after each simplication and should stop if expression turns determinate. In addition, we may use following results for rationalizing expressions involving surds : a b a b) a + b ) a /3 b /3 a b) a 2/3 + a /3 b /3 + b 2/3) * Version.: Sep 2, :56 am

2 OpenStax-CNX module: m Example Problem : Determine it Solution : nd it, x ) 2x ) x 2x 2 x Here, indeterminate form is 0/0. We simplify to change indeterminate form and x ) 2x ) 2x 2 x x ) 2x ) x ) 2x + ) This is determinate form. Plugging for x, we have : 2x ) x + ) 2x + ) L 6 Example 2 Problem : Determine it x 0x8 + x) 3 2x Solution : The indeterminate form is -. Simplifying, we have : We know that : f x) x) 3 2x8 + x) 3 a /3 b /3 a b) a 2/3 + a /3 b /3 + b 2/3) Using this identity : Substituting in the given expression, x) x) x) ) x8 + x) x) 3 This is a determinate form. Plugging 0 for x, Example 3 Problem : Determine it : L 2X8 3 x ) ) )

3 OpenStax-CNX module: m Solution : nd it, x2 ) x) x2 ) x) x 0 x Here, indeterminate form is 0/0. We simplify to change indeterminate form and x x 2 + x ) x{ x 2 ) + x)} x) { x 2 ) + x)} This simplied form is not indeterminate. Plugging 0 for x : L 2.2 Using standard it form There is an important algebraic form which is used as standard form. The standard form is n is rational number) : x n a n x 0 x a nan For rational n, the expansion of the expression in the it is given by : xn a n x a xn + ax n 2 + a 2 x n a n The expression on the right hand side of the equation is not indeterminate. Thus, it is obtained simply by plugging a for x : L a n + axa n 2 + a 2 a n a n L na n Example 4 Problem : Determine it x 3/2 a 3/2 x ax /2 a /2 Solution : Here, indeterminate form is 0/0. Substituting y x /2 and b a /2. As x > a, y > b. Using formula, x 3/2 a 3/2 x /2 a y3 b 3 /2 y b L 3b 2 3 a /2) 2 3a Example 5 Problem : Determine it

4 OpenStax-CNX module: m Solution : x + h) /n x /n h 0 h Here, indeterminate form is 0/0. Rearranging and using formulae, x + h) /n x /n h 0 x + h x As h 0, x + h x. Using formulae, L n x n.3 Canceling linear factors for rational function) If expression is a rational function, then it is likely that both numerator and denominator become zero at xa such that given expression has 0/0 indeterminate form. Clearly, then x-a) is a factor of both numerator and denominator. Canceling common linear factor, we get determinate form. We evaluate it by plugging iting value of x in the expression. Example 6 Problem : Determine it : x 3 3x 2 x + 3 x 3 x 2 x 6 Solution : Here, indeterminate form is 0/0. The numerator and the denominator individually tend to 0 as x 3. It means x-3) is factor of both numerator and denominator. Dividing polynomials long method or otherwise), we nd the quotient. We replace expression in terms of linear factor and quotient : x 3 3x 2 x + 3 x 2 x 3) x 2 ) x 6 x 3) x + 2) x 2 ) x + 2) This is determinate form. Plugging 3 for x, we have : L Limits of algebraic function when variable tends to innity In this case, we are required to estimate behavior of expression when variable approaches very large positive or negative value. The basic idea is to obtain each term in reciprocal form. As variable approaches innity, the reciprocal term approaches zero. There are two methods. However, we need to simplify expression before using either of these methods. We should also note that these two methods are completely equivalent. We can use either of two methods to evaluate its. Application of a method is a matter of choice and ease. : Dividing each term by highest power of variable

5 OpenStax-CNX module: m We divide each term of numerator and denominator by x raised to highest positive power in the given expression. Then, we use following it, c x, x n 0; n > 0 2: Take out highest power of variable We take out x raised to highest power from numerator and denominator separately. The highest power variable is considered separately for numerator and denominator. This is unlike previous case in which we use highest power variable of the whole expression. Finally, we evaluate resulting expression using it rules specied above for the reciprocals and using following additional its, When x, x n 0 if n < 0 x n if n 0 x n if n > 0 We should again emphasize that two methods are essentially equivalent. Radical and negative variable Dividing radical by variable poses diculty when variable represents negative value. Such is the case, when we are evaluating it in which variable is approaching negative innity. Clearly, variable has negative value in such cases. We use following rule : If x < 0, then x x 2 ) In order to understand working of this rule, let us consider a radical : x2 + 4x) We are required to divide the radical by x, when it is known to be negative. Following the fact stated above, x2 ) x2 + 4x) + 4x x x ) x Example 7 Problem : Determine it : x 4 + 2x x 2x 4 x + 2 Solution : Here, indeterminate form is /. Dividing each term by x 4 x 4 + 2x x 4 x x + 3 x 4 2 x x 4 This is determinate form. As x, 2/x 0, 3 x 4 0, Numerator and as x, x 3 0, 2 x 4 0, Denominator 2. Hence, it is : L 2

6 OpenStax-CNX module: m Alternatively, we can employ second method to evaluate it. Taking out x 4 x 4 + 2x x 4 x + 2 x4 + 2 x + ) 3 x 4 x ) 4 2 x x 4 Example 8 Problem : Determine it : L 2 a n + b n a n b n ; a > b > x Solution : Here, indeterminate form is /. By inspection, we see that a/b > and b/a <. As we know x,c x 0 if c <. Hence, we are required to get terms in the form b/a raised to some power. Dividing numerator and denominator by a n, we have : a n + b n a n b n + b n a) ) b n This is determinate form. As n, b/a) n 0. Hence, Example 9 Problem : Determine it : 2 L x n n n n2 n 3 Solution : The indeterminate form is /. Writing expression of sum of square of natural numbers, we have : 2 n n n n2 n n 2 n 3 As n, 3/n, /n 2 0 Example 0 Problem : Determine it : n n + ) 2n + ) n + ) 2n + ) 2n 3 2n 2 L 2 2 x { x 2 + 4x) x 2 4x)} a n + n 2 ) 2 Solution : Here, indeterminate form is -. Rationalizing, we have : x2 + 4x) x 2 8x 4x) { x 2 + 4x) + x 2 4x)} Dividing by x. Note x is negative. Hence x x 2 )

7 OpenStax-CNX module: m As x, 4 x ) { + 4 x 4 x) } L 4 3 Exercises Exercise Solution on p. 9.) Determine it x x 2 2 x 4 Exercise 2 Solution on p. 9.) Determine it x{ x + c) x} x Exercise 3 Solution on p. 9.) Determine it {x x 2 + x)} x Exercise 4 Solution on p. 9.) Determine it : x { ) + x x} x Exercise 5 Solution on p. 0.) Determine it x 5/2 a 5/2 x ax /2 a /2 Exercise 6 Solution on p. 0.) Determine it x x 2 x 2 Exercise 7 Solution on p. 0.) Determine it : x p + x p + x q + x q + ; p > 0, q > 0 x Exercise 8 Solution on p..) Determine it :

8 OpenStax-CNX module: m x) 3 x 0 3x + 2x 2 Exercise 9 Solution on p..) Determine it : x 2) + x 2 x 2 x2 4) Exercise 0 Solution on p..) Determine it : x x x2 ) + x ) Exercise Solution on p. 2.) Determine it x /6 2 x /3 4 x 64 Exercise 2 Solution on p. 2.) Determine it : { x 2 + 8) 0 x 2 )} x { x 2 + 3) 5 x 2 )} Exercise 3 Solution on p. 2.) Determine it : x 7 2x 5 + x x 3 3x Exercise 4 Solution on p. 3.) Determine it : x a a + 2x) 3x) 3a + x) 2 x ; x a

9 OpenStax-CNX module: m Solutions to Exercises in this Module Solution to Exercise p. 7) Here, indeterminate form is -. We simplify to change the form of expression from determinate, x 2 2 x 4 x 2 ) 2 x 2 ) x 2 + ) x x 2 ) x 2 + ) x 2 x 2 ) x 2 + ) x 2 + ) This form is determinate. Plugging for x, we have : L 2 Solution to Exercise p. 7) Here, indeterminate form is -. Rationalizing, we have : Dividing by x, As x, c/x 0 c x x{ x + c) x} { x + c) + x} c { + c x) + } L c 2 Solution to Exercise p. 7) Here, indeterminate form is -. Rationalizing surd, we have : Dividing each of terms by x, we have : {x x 2 + x)} {x x 2 + x)}{x + x 2 + x)} {x + x 2 + x)} x 2 x 2 x ) {x + x 2 + x)} x) {x + x 2 + x)} This is determinate form. As x->, /x -> 0. ) { + + x) } L + 2 Solution to Exercise p. 7) Here, indeterminate form is -. Rationalizing surds, we have :

10 OpenStax-CNX module: m x ) { x + x) x}{ x + x) + x} { + x x} { x + x) + x} x + x x) x { x + x) + x} { x + x) + x} Dividing numerator and denominator by x, This is determinate form. As x, x 0, { + x ) + } L 2 Solution to Exercise p. 7) Here, indeterminate form is 0/0. We put y x /2, b a /2 ; x a, y b. Using formulae : x 5/2 a 5/2 x /2 a x5 a 5 /2 x a L 5b 4 L 5 a /2) 4 L 5a 2 Solution to Exercise p. 7) Here, indeterminate form is -. We simplify to change indeterminate form and nd it, x 2 x 2 x 2 x ) x + ) x + 2 x ) x + ) x x ) x + ) x + ) It is determinate form. Plugging for x, we have : L 2 Solution to Exercise p. 7) Here, indeterminate form is /. We divide each term by x p. x p + x p + x q + x q + xp + x + ) x p x q + x + ) x q As x, x and x 0, and p

11 OpenStax-CNX module: m7542 x p q ; if p > q x p q ; if p q Hence, x p q 0; if p < q L ; if p > q L ; if p q L 0; if p < q Solution to Exercise p. 7) Here, indeterminate form is 0/0. Using standard form, This is determinate form. x 0, + x + x) 3 3x + 2x 2 + x)3 + x) X x 3x + 2x 2 + x)3 + x) X 3 + 2x L 3X2 3 Solution to Exercise p. 8) Here, indeterminate form is 0/0. We simplify to change indeterminate form and nd it, x 2) + x 2 x 2) x 2 x2 4) x2 4) + x2 4) ) ) x 2 x ) x + 2) x + 2 x2 4) x 2) + ) x + 2) x + 2 x + 2) This is determinate form. Plugging 2 for x, we have : L 2 Solution to Exercise p. 8) Here, indeterminate form is 0/0. Rationalizing surds, Each term its to na n. x x2 ) + x ) x ) x ) X x ) { x2 ) x ) X x )} + { x ) x ) X x )}

12 OpenStax-CNX module: m This is determinate form. L Solution to Exercise p. 8) Here, indeterminate form is 0/0. Using standard formulae, we have : It is determinate form. Evaluating, we have : x /6 2 x /3 4 x/6 64 /6 x /3 64 /3 L x /6 64 /6 x 64 x /3 64 /3 x 64 6 X64/6 3 X64/3 L 4 Solution to Exercise p. 8) Here, indeterminate form is -. Rationalizing surds, { x 2 + 8) 0 x 2 )} { x 2 + 3) 5 x 2 )} { x2 + 8) 0 x 2 )}{ x 2 + 8) + 0 x 2 )}{ x 2 + 3) + 5 x 2 )} { x 2 + 8) + 0 x 2 )}{ x 2 + 3) 5 x 2 )}{ x 2 + 3) + 5 x 2 )} x x 2) { x 2 + 3) + 5 x 2 )} x x 2 ) { x 2 + 8) + 0 x 2 )} 2x 2 2 ) { x 2 + 3) + 5 x 2 )} 2x 2 2) { x 2 + 8) + 0 x 2 )} { x 2 + 3) + 5 x 2 )} { x 2 + 8) + 0 x 2 )} This is in determinate form. Plugging for x, we have : L Solution to Exercise p. 8) Here, indeterminate form is 0/0. The numerator and denominator tend to 0 as x- >. It means x-) is factor of both numerator and denominator. Dividing polynomials long method or otherwise) and using quotient : x 7 2x 5 + x ) x 6 + x 5 x 4 x 3 x 2 x ) x 3 3x x ) x 2 2x 2) x 6 + x 5 x 4 x 3 x 2 x ) x 2 2x 2) This is determinate form. Plugging by x, we know :

13 OpenStax-CNX module: m L 3 3 Solution to Exercise p. 8) Indeterminate form is 0/0. We simplify the expression to change indeterminate form and nd it, a + 2x) 3x) 3a + x) 2 x { a + 2x) 3x)}{ a + 2x) + 3x)}{ 3a + x) + 2 x} { a + 2x) + 3x)}{ 3a + x) 2 x}{ 3a + x) + 2 x} a + 2x 3x) { 3a + x) + 2 x} 3a + x 4x) { a + 2x) + 3x)} a x) { 3a + x) + 2 x} 3 a x) { a + 2x) + 3x)} { 3a + x) + 2 x} 3{ a + 2x) + 3x)} This is not in indeterminate form. Plugging a for x {2 a + 2 a} L 3{ 3a) + 3a)} {4 a} 6 3a) L