Solved Examples. (Highest power of x in numerator and denominator is ½. Dividing numerator and denominator by x)

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1 Solved Examples Example 1: (i) (ii) lim x (x 4 + 2x 3 +3) / (2x 4 -x+2) lim x x ( (x+c)- x) (iii) lim n ( (2n-1)-2n)/ (n 2 +1) (iv) lim x 0 ((1+x) 5-1)/3x+5x 2 (v) lim x 2 ( (x+7)-3 (2x-3))/((x+6) 1/3-2 (3x-5) 1/3 ) Solution: (i) If we divide the polynomial in numerator and denominator by x 4 (as x 4 is highest power of x in denominator and numerator. => lim x (1+2/x+3/x 4 )/(2-1/x 3 +2/x 4 ) = 1/2 (ii) lim x x( (x+c)- x) Rationalize the expression by multiplying and dividing by ( (x+c)+ x) => lim x x ( (x+c)- x)( (x+c)+ x) /( (x+c)+ x) => lim x ( x (x+c-x))/( (x+c)+ x) (Highest power of x in numerator and denominator is ½. Dividing numerator and denominator by x) = lim x c /(1+ (1+c/x)) c/2 (iii) lim n ( (2n-1)-2n)/ (n 2 +1) Observe that 1-2 = = = -1, and so on.

2 In numerator, each group of two terms is reduced to -1, and there are n groups. Now the expressions becomes, lim n (-n)/ (n 2 +1) = lim n (-1)/ (1+1/n 2 ) = (-1)/ (1+0) = -1 (iv) lim x ((1+x) 5-1)/(3x+5x 2 ) (iv) lim x ((1+x) 5-1)/(3x+5x 2 ) Method 1: On expanding (1 + x) 5 we get = 5/3 Method 2: lim x 0 ((1+x) )/((1+x)-1). x/(3x+5x 2 ) = /(3+0) = 5/3

3 Example 2: (i) lim x (x+sin x+1)/(x+cos x ) (ii) lim x cos{log((x-1)/x) } (iii) lim x ((3x 2 +2x 2 )sin(1/x)+ x 3 +5)/( x 3 + x 2 + x +1)

4 (iv) lim x 0 (sin -1 x - tan -1 x)/x 3 (v) lim x a (x sin α sin x)/(x-α) Solution: (i) lim x x(e 1/x - e -1/x ) => 1/x -> 0 let h = 1/x Method 1: = lim n (1+1/2n) n n->, 1/n ->0 Let h = 1/n

5 = e 1/2 Methods 2: We can use the result lim x 0 (1+1/x) x =e directly L = lim h 0 ((1+h/2) 2/h ) 1/2 = e 1/2 (iii) Let us do some analysis here

6 Let g(x) = x 2 + 5/2 So as x->, g(x) -> Hence lim x (1-1/g(x)) g(x) is very similar to the expression lim x (1-1/x) x and so we can equate this to the expression of e -1. So the given limit is simplified to finding limit of x-> e = e -8 (iv) (v) Method I:

7 Do these problems using L' Hospital's Rule, if you know, otherwise we shall study this rule in next chapter. Method 2: lim x 1 x cot Πx... here put x = (1 + h) cot(π+π h) = lim h 0 (1+h) = lim h 0 [(1+h) 1/h h/tan Πh ] Πh/tan Πh. (1/Π) = lim h 0 e = e 1/Π Note: cot (Π+Πh) = cot Πh = 1/tan Πh Example 4: (i) lim n (cos x/2 cos x/4 cos x/8...cos x/2 n ) (ii) lim n sin 2n x (iii) (1+ cos Πx) cot 2 Πx Solution: (i) Let P = cos x/2.cos x/x 2.cos x/2 3...cos x/2 n => P sin x/2 n = cos(x/2).cos(x/2 2 ).cos(x/2 3 )...cos(x/2 n ).sin(x/2 n )

8 = 1/2.(cos Π/2)(cos x/2 2 )(cos x/2 3 )...cos x/2 (n-1).sin x/2 (n-1) = 1/2 2.(cos x/2)(cos x/2 2 )(cos x/2 3 )...cos x/2 (n-2).sin x/2 (n-2) Proceeding similarly we get P.sin x/2 n =1/2 n-1.cos(x/2)sin x/2 n-(n-1) = 1/2 n-1 cos x/2.sin x/2 = sin x/2 n => P = sin x/(2 n sin(x/2 n ) ) => lim n Π P = lim n Π {sin x/(2 n sin(x/2 n ) )} = lim n Π sinx/x ((x/2 n )/sin x/2 n ) = sinx/x (ii) L = lim n Π sin2 n x At sin x = 0; i.e. x = np L = lim n Π (0) 2n =0 At sin x = (±1) i.e. x = m p/2 (m is odd integer) L= lim n Π (±1) 2n =1 For other values of x we have 0<sin 2 x <1 L= lim n Π (sin 2n x)=0 lim x (2l+1)Π/2, n-> (sinl) 2 n = Not defined. (indeterminated) Where l I Note: (1.1) 1 = 1.1 (1.1) 2 = 1.21

9 (1.1) 3 = (1.1) 3 = 1.03 Thus we observe that lim x 1 +, y xy approaches one if x->1 + faster than y->. And lim x 1 +, y xy approaches infinity if y-> faster than x->1 +.

Math 106 Fall 2014 Exam 2.1 October 31, ln(x) x 3 dx = 1. 2 x 2 ln(x) + = 1 2 x 2 ln(x) + 1. = 1 2 x 2 ln(x) 1 4 x 2 + C

Math 106 Fall 2014 Exam 2.1 October 31, ln(x) x 3 dx = 1. 2 x 2 ln(x) + = 1 2 x 2 ln(x) + 1. = 1 2 x 2 ln(x) 1 4 x 2 + C Math 6 Fall 4 Exam. October 3, 4. The following questions have to do with the integral (a) Evaluate dx. Use integration by parts (x 3 dx = ) ( dx = ) x3 x dx = x x () dx = x + x x dx = x + x 3 dx dx =