Question 1

What is _ x 1 x^ (n^2-1) -1 x^ (n+1) -1 equal to, where n>1 is a natural number ?

Accepted Answer

The given limit is _ x 1 x^ (n^2-1) -1 x^ (n+1) -1 . As x 1, the expression takes the indeterminate form 0 0 . Applying L'Hospital's rule, we differentiate the numerator and the denominator with respect to x: _ x 1 (n^2-1)x^ n^2-2 (n+1)x^n Substituting x=1, we get: n^2-1 n+1 (n-1)(n+1) n+1 n-1 Answer: n-1

Question 2

What is _ x 0 10^ x -1 x equal to ?

Accepted Answer

The given limit is _ x 0 10^ x -1 x Multiplying and dividing the expression by x: _ x 0 ( 10^ x -1 x x x ) _ x 0 ( 10^ x -1 x x ) Using the standard limit _ t 0 a^t-1 t = a, where t = x 0 as x 0: 10 0 10 1 = 10 Answer: 10

Question 3

Consider the following for the two (02) items that follow: Let the function f(x) = x^2 + 9. What is _ x 0 f(x) - 3 f(x)+7 - 4 equal to?

Accepted Answer

Given f(x) = x^2 + 9. The limit is _ x 0 x^2 + 9 - 3 x^2 + 16 - 4 . Rationalizing the numerator and the denominator: _ x 0 ( x^2 + 9 - 3)( x^2 + 9 + 3) ( x^2 + 16 - 4)( x^2 + 16 + 4) x^2 + 16 + 4 x^2 + 9 + 3 _ x 0 x^2 + 9 - 9 x^2 + 16 - 16 x^2 + 16 + 4 x^2 + 9 + 3 _ x 0 x^2 x^2 x^2 + 16 + 4 x^2 + 9 + 3 Cancelling x^2 and substituting x = 0: 16 + 4 9 + 3 = 4 + 4 3 + 3 = 8 6 = 4 3 Answer: 4 3

Question 4

For the following three (03) items: Consider the function f(x) = x|x|. What is _ x -1 f(x) equal to?

Accepted Answer

Given the function f(x) = x|x|. We need to evaluate _ x -1 f(x). Since x approaches -1, x is negative, so |x| = -x. Thus, f(x) = x(-x) = -x^2 for x _ x -1 f(x) = _ x -1 (-x^2) = -(-1)^2 = -1. Answer: -1

Question 5

For the following two (02) items: Let f(x) = cases 1 - 2x x^2 , & x 0 cases What is _ x 0^- f(x) equal to?

Accepted Answer

For x To find the left-hand limit, we evaluate _ x 0^- f(x) _ x 0^- 1 - 2x x^2 Using the trigonometric identity 1 - 2x = 2 ^2 x _ x 0^- 2 ^2 x x^2 = 2 ( _ x 0^- x x )^2 Since _ x 0 x x = 1, we get: 2 (1)^2 = 2 Answer: 2

Question 6

For the following two (02) items: Let f(x) = cases 1 - 2x x^2 , & x 0 cases What is _ x 0^+ f(x) equal to?

Accepted Answer

For x > 0, the function is given by f(x) = x 16 + x - 4 To find the right-hand limit, we evaluate _ x 0^+ f(x) _ x 0^+ x 16 + x - 4 Rationalizing the denominator by multiplying the numerator and the denominator by 16 + x + 4: _ x 0^+ x ( 16 + x + 4) ( 16 + x - 4)( 16 + x + 4) _ x 0^+ x ( 16 + x + 4) 16 + x - 16 _ x 0^+ x ( 16 + x + 4) x Canceling x from the numerator and the denominator: _ x 0^+ ( 16 + x + 4) Substituting x = 0: 16 + 0 + 4 = 4 + 4 = 8 Answer: 8

Question 7

Consider the following for the two (02) items that follow: Let the function f(x) = [x], where [ ] is the greatest integer function and g(x) = |x|. What is _ x 0 \ f(x) g(x)\ equal to?

Accepted Answer

Given f(x) = [x] and g(x) = |x|. To find _ x 0 f(x) g(x), we evaluate the left-hand limit (LHL) and the right-hand limit (RHL). For LHL, as x 0^ - , [x] = -1 and |x| = -x. _ x 0^ - f(x) g(x) = _ x 0^ - ( (-1))(-x) = (- 1)(0) = 0 For RHL, as x 0^ + , [x] = 0 and |x| = x. _ x 0^ + f(x) g(x) = _ x 0^ + ( 0)(x) = (0)(0) = 0 Since LHL = RHL = 0, the limit exists and is equal to 0. Answer: 0

Question 8

Consider the following for the two (02) items that follow: Let the function f(x) = [x], where [ ] is the greatest integer function and g(x) = |x|. What is _ x 0 f(x) g(x) equal to?

Accepted Answer

The given functions are f(x) = [x] and g(x) = |x|. To find _ x 0 f(x) g(x) , we evaluate the left-hand limit (LHL) and right-hand limit (RHL) at x = 0. For the right-hand limit (x 0^+): When 0 _ x 0^+ [x] |x| = _ x 0^+ 0 x = _ x 0^+ 0 x = 0 For the left-hand limit (x 0^-): When -1 _ x 0^- [x] |x| = _ x 0^- (-1) -x = _ x 0^- - 1 -x = _ x 0^- 1 x = - Since the left-hand limit and right-hand limit are not equal, the limit does not exist.

Question 9

Consider the following for the two (02) items that follow: Let the function f(x) = x^2 - 1. What is _ x 1 \ f f(x)\ equal to?

Accepted Answer

Given f(x) = x^2 - 1. The composite function f f(x) is given by: f(f(x)) = f(x^2 - 1) = (x^2 - 1)^2 - 1 Since f(f(x)) is a polynomial and continuous everywhere, the limit can be evaluated by direct substitution: _ x 1 \ f f(x)\ = f(f(1)) f(1) = 1^2 - 1 = 0 f(f(1)) = f(0) = 0^2 - 1 = -1 Thus, the limit is -1. Answer: -1

Limits — Previous Year Questions & Solutions