QBManager

Limits, Continuity and Differentiability

79484 The value of $lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} =$

1

\log 5

2 0

3 1

4

2 \log 5

Explanation:

(A) : Given,
$lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x}$
From L- Hospital's's rule-
$= lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} = lim_{x \to 0} \frac{5^{x} \log 5 - 5^{- x} \log 5 \times (- 1)}{2}$
$= lim_{x \to 0} \frac{5^{x} \log 5 + 5^{- x} \log 5}{2} = \frac{\log 5 + \log 5}{2} = \log 5$

Karnataka CET-2006

Limits, Continuity and Differentiability

79486 $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} =$

1 1

2 -1

3 zero

4

- 1 / 2

Explanation:

(D) : Given, $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}}$
Applying L'-Hospital's rule,
$lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} = lim_{x \to 1} \frac{\frac{1}{x} - 1}{- 2 + 2 x}$
$lim_{x \to 1} \frac{(1 - x)}{2 x (x - 1)} = lim_{x \to 1} \frac{- (x - 1)}{2 x (x - 1)}$
$lim_{x \to 1} (- \frac{1}{2 x}) = - \frac{1}{2}$

Karnataka CET-2000

Limits, Continuity and Differentiability

79487 If $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$ , then :

1

a = 1

and

b = 1

2

a = 1

and

b = - 1

3

a = 1

and

b = - 2

4

a = 1

and

b = 2

Explanation:

(C) : Given, $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$
$lim_{x \to \infty} [\frac{x^{3} + 1 - a x^{3} - b x^{2} - a x - b}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{x^{3} (1 - a) - b x^{2} - a x - b + 1}{x^{2} + 1}] = 2$
For the infinite limit exist, the coefficient of $x^{3}$ must be zero.
$∴ 1 - a = 0$
$a = 1$
$lim_{x \to \infty} [\frac{- b x^{2} - x - b + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1) - x + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1)}{x^{2} + 1} - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [- b - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$- b - 0 + 0 = 2$
$b = - 2$

Karnataka CET-2000

Limits, Continuity and Differentiability

79488 If $f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$ , then $lim_{x \to 0} \frac{f (x)}{x^{2}}$ is :

1 -1

2 3

3 1

4 zero

Explanation:

(C) $: Given, f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$
$f (x) = \sin x (x^{2} - x) - \cos x \cdot (x^{3} - 2 x^{2}) + \tan x (x^{3} - 2 x^{3})$
$f (x) = x (x - 1) \sin x - x^{2} (x - 2) \cos x - x^{3} \tan x$
$\frac{f (x)}{x^{2}} = \frac{(x - 1) \sin x}{x} - (x - 2) \cos x - x \tan x$
$= \sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = lim_{x \to 0} [\sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x]$
$= \sin 0 - 1 - (0 - 2) \cos 0^{\circ} - 0 = 0 - 1 + 2$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = 1$

Karnataka CET-2002

Limits, Continuity and Differentiability

79484 The value of $lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} =$

1

\log 5

2 0

3 1

4

2 \log 5

Explanation:

(A) : Given,
$lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x}$
From L- Hospital's's rule-
$= lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} = lim_{x \to 0} \frac{5^{x} \log 5 - 5^{- x} \log 5 \times (- 1)}{2}$
$= lim_{x \to 0} \frac{5^{x} \log 5 + 5^{- x} \log 5}{2} = \frac{\log 5 + \log 5}{2} = \log 5$

Karnataka CET-2006

Limits, Continuity and Differentiability

79486 $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} =$

1 1

2 -1

3 zero

4

- 1 / 2

Explanation:

(D) : Given, $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}}$
Applying L'-Hospital's rule,
$lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} = lim_{x \to 1} \frac{\frac{1}{x} - 1}{- 2 + 2 x}$
$lim_{x \to 1} \frac{(1 - x)}{2 x (x - 1)} = lim_{x \to 1} \frac{- (x - 1)}{2 x (x - 1)}$
$lim_{x \to 1} (- \frac{1}{2 x}) = - \frac{1}{2}$

Karnataka CET-2000

Limits, Continuity and Differentiability

79487 If $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$ , then :

1

a = 1

and

b = 1

2

a = 1

and

b = - 1

3

a = 1

and

b = - 2

4

a = 1

and

b = 2

Explanation:

(C) : Given, $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$
$lim_{x \to \infty} [\frac{x^{3} + 1 - a x^{3} - b x^{2} - a x - b}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{x^{3} (1 - a) - b x^{2} - a x - b + 1}{x^{2} + 1}] = 2$
For the infinite limit exist, the coefficient of $x^{3}$ must be zero.
$∴ 1 - a = 0$
$a = 1$
$lim_{x \to \infty} [\frac{- b x^{2} - x - b + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1) - x + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1)}{x^{2} + 1} - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [- b - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$- b - 0 + 0 = 2$
$b = - 2$

Karnataka CET-2000

Limits, Continuity and Differentiability

79488 If $f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$ , then $lim_{x \to 0} \frac{f (x)}{x^{2}}$ is :

1 -1

2 3

3 1

4 zero

Explanation:

(C) $: Given, f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$
$f (x) = \sin x (x^{2} - x) - \cos x \cdot (x^{3} - 2 x^{2}) + \tan x (x^{3} - 2 x^{3})$
$f (x) = x (x - 1) \sin x - x^{2} (x - 2) \cos x - x^{3} \tan x$
$\frac{f (x)}{x^{2}} = \frac{(x - 1) \sin x}{x} - (x - 2) \cos x - x \tan x$
$= \sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = lim_{x \to 0} [\sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x]$
$= \sin 0 - 1 - (0 - 2) \cos 0^{\circ} - 0 = 0 - 1 + 2$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = 1$

Karnataka CET-2002

Limits, Continuity and Differentiability

79484 The value of $lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} =$

1

\log 5

2 0

3 1

4

2 \log 5

Explanation:

(A) : Given,
$lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x}$
From L- Hospital's's rule-
$= lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} = lim_{x \to 0} \frac{5^{x} \log 5 - 5^{- x} \log 5 \times (- 1)}{2}$
$= lim_{x \to 0} \frac{5^{x} \log 5 + 5^{- x} \log 5}{2} = \frac{\log 5 + \log 5}{2} = \log 5$

Karnataka CET-2006

Limits, Continuity and Differentiability

79486 $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} =$

1 1

2 -1

3 zero

4

- 1 / 2

Explanation:

(D) : Given, $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}}$
Applying L'-Hospital's rule,
$lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} = lim_{x \to 1} \frac{\frac{1}{x} - 1}{- 2 + 2 x}$
$lim_{x \to 1} \frac{(1 - x)}{2 x (x - 1)} = lim_{x \to 1} \frac{- (x - 1)}{2 x (x - 1)}$
$lim_{x \to 1} (- \frac{1}{2 x}) = - \frac{1}{2}$

Karnataka CET-2000

Limits, Continuity and Differentiability

79487 If $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$ , then :

1

a = 1

and

b = 1

2

a = 1

and

b = - 1

3

a = 1

and

b = - 2

4

a = 1

and

b = 2

Explanation:

(C) : Given, $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$
$lim_{x \to \infty} [\frac{x^{3} + 1 - a x^{3} - b x^{2} - a x - b}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{x^{3} (1 - a) - b x^{2} - a x - b + 1}{x^{2} + 1}] = 2$
For the infinite limit exist, the coefficient of $x^{3}$ must be zero.
$∴ 1 - a = 0$
$a = 1$
$lim_{x \to \infty} [\frac{- b x^{2} - x - b + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1) - x + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1)}{x^{2} + 1} - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [- b - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$- b - 0 + 0 = 2$
$b = - 2$

Karnataka CET-2000

Limits, Continuity and Differentiability

79488 If $f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$ , then $lim_{x \to 0} \frac{f (x)}{x^{2}}$ is :

1 -1

2 3

3 1

4 zero

Explanation:

(C) $: Given, f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$
$f (x) = \sin x (x^{2} - x) - \cos x \cdot (x^{3} - 2 x^{2}) + \tan x (x^{3} - 2 x^{3})$
$f (x) = x (x - 1) \sin x - x^{2} (x - 2) \cos x - x^{3} \tan x$
$\frac{f (x)}{x^{2}} = \frac{(x - 1) \sin x}{x} - (x - 2) \cos x - x \tan x$
$= \sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = lim_{x \to 0} [\sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x]$
$= \sin 0 - 1 - (0 - 2) \cos 0^{\circ} - 0 = 0 - 1 + 2$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = 1$

Karnataka CET-2002

Limits, Continuity and Differentiability

79484 The value of $lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} =$

1

\log 5

2 0

3 1

4

2 \log 5

Explanation:

(A) : Given,
$lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x}$
From L- Hospital's's rule-
$= lim_{x \to 0} \frac{5^{x} - 5^{- x}}{2 x} = lim_{x \to 0} \frac{5^{x} \log 5 - 5^{- x} \log 5 \times (- 1)}{2}$
$= lim_{x \to 0} \frac{5^{x} \log 5 + 5^{- x} \log 5}{2} = \frac{\log 5 + \log 5}{2} = \log 5$

Karnataka CET-2006

Limits, Continuity and Differentiability

79486 $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} =$

1 1

2 -1

3 zero

4

- 1 / 2

Explanation:

(D) : Given, $lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}}$
Applying L'-Hospital's rule,
$lim_{x \to 1} \frac{1 + \log x - x}{1 - 2 x + x^{2}} = lim_{x \to 1} \frac{\frac{1}{x} - 1}{- 2 + 2 x}$
$lim_{x \to 1} \frac{(1 - x)}{2 x (x - 1)} = lim_{x \to 1} \frac{- (x - 1)}{2 x (x - 1)}$
$lim_{x \to 1} (- \frac{1}{2 x}) = - \frac{1}{2}$

Karnataka CET-2000

Limits, Continuity and Differentiability

79487 If $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$ , then :

1

a = 1

and

b = 1

2

a = 1

and

b = - 1

3

a = 1

and

b = - 2

4

a = 1

and

b = 2

Explanation:

(C) : Given, $lim_{x \to \infty} [\frac{x^{3} + 1}{x^{2} + 1} - (a x + b)] = 2$
$lim_{x \to \infty} [\frac{x^{3} + 1 - a x^{3} - b x^{2} - a x - b}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{x^{3} (1 - a) - b x^{2} - a x - b + 1}{x^{2} + 1}] = 2$
For the infinite limit exist, the coefficient of $x^{3}$ must be zero.
$∴ 1 - a = 0$
$a = 1$
$lim_{x \to \infty} [\frac{- b x^{2} - x - b + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1) - x + 1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [\frac{- b (x^{2} + 1)}{x^{2} + 1} - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$lim_{x \to \infty} [- b - \frac{x}{x^{2} + 1} + \frac{1}{x^{2} + 1}] = 2$
$- b - 0 + 0 = 2$
$b = - 2$

Karnataka CET-2000

Limits, Continuity and Differentiability

79488 If $f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$ , then $lim_{x \to 0} \frac{f (x)}{x^{2}}$ is :

1 -1

2 3

3 1

4 zero

Explanation:

(C) $: Given, f (x) = | \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & 1 \end{array} |$
$f (x) = \sin x (x^{2} - x) - \cos x \cdot (x^{3} - 2 x^{2}) + \tan x (x^{3} - 2 x^{3})$
$f (x) = x (x - 1) \sin x - x^{2} (x - 2) \cos x - x^{3} \tan x$
$\frac{f (x)}{x^{2}} = \frac{(x - 1) \sin x}{x} - (x - 2) \cos x - x \tan x$
$= \sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = lim_{x \to 0} [\sin x - \frac{\sin x}{x} - (x - 2) \cos x - x \tan x]$
$= \sin 0 - 1 - (0 - 2) \cos 0^{\circ} - 0 = 0 - 1 + 2$
$lim_{x \to 0} \frac{f (x)}{x^{2}} = 1$

Karnataka CET-2002