QBManager

Limits, Continuity and Differentiability

79474 $\sum_{r = 1}^{n} (2 r - 1) = x$ , then
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots . . + \frac{n^{3}}{x^{2}}] =$

1

\frac{1}{2}

2 1

3

\frac{1}{4}

4 4

Explanation:

(C) : Given, $\sum_{r = 1}^{n} (2 r - 1) = x$
$2 \sum_{r = 1}^{n} r - \sum 1 = x$ s
$2 [\frac{n (n + 1)}{2}] - n$
$[\begin{matrix} 2 (n + 1) \\ x_{4} \end{matrix}] - n = x \Rightarrow n^{2} + n - n = x \Rightarrow n^{2} = x$
$x = n$
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots \dots . . + \frac{n^{3}}{x^{2}}]$
$= lim_{n \to 0} [\frac{1}{x^{2}} (1^{3} + 2^{3} + 3^{3} + \dots \dots \dots + n^{3})]$
$= lim_{n \to 0} [\frac{1}{n^{4}} \times {\frac{n (n + 1)}{2}}^{2}] = \frac{1}{4} lim_{n \to 0} [\frac{(n + 1)^{2}}{n^{2}}]$
$= \frac{1}{4} lim_{n \to 0} [\frac{n^{2} + 1 + 2 n}{n^{2}}] = \frac{1}{4} lim_{n \to 0} [1 + \frac{1}{n^{n}} + \frac{2}{n}]$
$= \frac{1}{4} [1 + \frac{1}{0} + \frac{2}{0}] = \frac{1}{4}$

Karnataka CET-2019

Limits, Continuity and Differentiability

79475 The value of $lim_{x \to 0} \frac{| x |}{x}$ is

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{| x |}{x}$
LHL $= lim_{h \to 0^{-}} \frac{| 0 - h |}{0 - h} \Rightarrow lim_{h \to 0^{-}} \frac{h}{- h} = - 1$
RHL $= lim_{h \to 0^{+}} \frac{| 0 + h |}{0 + h} \Rightarrow lim_{h \to 0^{+}} \frac{h}{h} = 1$
So, LHL $\neq$ RHL
$∴$ Limit does not exist at $x = 0$

Karnataka CET-2018

Limits, Continuity and Differentiability

79476 If the function $f (x)$ satisfies $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$ , then $lim_{x \to 1} f (x) =$

1 1

2 2

3 0

4 3

Explanation:

(B) : Given, $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$
$\frac{lim_{x \to 1} (f (x) - 2)}{lim_{x \to 1} x^{2} - 1} = π$
${lim}_{x \to 1}^{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$
$lim_{x \to 1} f (x) - lim_{x \to 1} 2 = π (1 - 1)$
$lim_{x \to 1} f (x) - 2 = 0$
$lim_{x \to 1} f (x) = 2$

Karnataka CET-2014

Limits, Continuity and Differentiability

79480 $lim_{n \to \infty} {n \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} =$

1

\frac{2 π}{3}

2

\frac{π}{6}

3

\frac{π}{3}

4 1

Explanation:

(A) : Given, $lim_{n \to \infty} {n \cdot \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {n \cdot 2 \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} = \frac{1}{2} lim_{n \to \infty} {n \cdot \sin \frac{4 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {\frac{n \cdot \sin \frac{4 π}{3 n}}{\frac{4 π}{3 n}} \times \frac{4 π}{3 n}} = \frac{4 π}{6} lim_{n \to \infty} {\frac{\sin 4 \frac{π}{3 n}}{\frac{4 π}{3 n}}}$
$∵ lim_{x \to \infty} \frac{\sin x}{x} = 1]$
$= \frac{4 π}{6} \times 1 = \frac{2 π}{3}$

Limits, Continuity and Differentiability

79481 $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$

1

\frac{2}{3}

2

\frac{2}{\sqrt{3}}

3

\frac{3 \sqrt{3}}{2}

4

\frac{2}{3 \sqrt{3}}

Explanation:

(D) : Given, $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$
$lim_{x \to a} \sqrt{a + 2 x} - \sqrt{3 x} \cdot \frac{\sqrt{a + 2 x} + \sqrt{3 x}}{\sqrt{a + 2 x} + \sqrt{3 x}} \cdot \frac{1}{\sqrt{3 a + x} - 2 \sqrt{x}} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{3 a + x} + 2 \sqrt{x}}$
$= lim_{x \to a} \frac{[(a + 2 x) - 3 x]}{[(3 a + x) - 4 x]} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{(a - x)}{3 (a - x)} \times \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{1}{3} \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= \frac{1}{3} \times \frac{\sqrt{3 a + a} + 2 \sqrt{a}}{\sqrt{a + 2 a} + \sqrt{3 a}} = \frac{1}{3} \times \frac{\sqrt{4 a} + 2 \sqrt{a}}{\sqrt{3 a} + \sqrt{3 a}}$
$= \frac{1}{3} \times \frac{2 \sqrt{a} + 2 \sqrt{a}}{2 \sqrt{3 a}} = \frac{1}{3} \times \frac{4 \sqrt{a}}{2 \sqrt{3 a}} = \frac{2}{3 \sqrt{3}}$

Karnataka CET-2011

Limits, Continuity and Differentiability

79474 $\sum_{r = 1}^{n} (2 r - 1) = x$ , then
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots . . + \frac{n^{3}}{x^{2}}] =$

1

\frac{1}{2}

2 1

3

\frac{1}{4}

4 4

Explanation:

(C) : Given, $\sum_{r = 1}^{n} (2 r - 1) = x$
$2 \sum_{r = 1}^{n} r - \sum 1 = x$ s
$2 [\frac{n (n + 1)}{2}] - n$
$[\begin{matrix} 2 (n + 1) \\ x_{4} \end{matrix}] - n = x \Rightarrow n^{2} + n - n = x \Rightarrow n^{2} = x$
$x = n$
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots \dots . . + \frac{n^{3}}{x^{2}}]$
$= lim_{n \to 0} [\frac{1}{x^{2}} (1^{3} + 2^{3} + 3^{3} + \dots \dots \dots + n^{3})]$
$= lim_{n \to 0} [\frac{1}{n^{4}} \times {\frac{n (n + 1)}{2}}^{2}] = \frac{1}{4} lim_{n \to 0} [\frac{(n + 1)^{2}}{n^{2}}]$
$= \frac{1}{4} lim_{n \to 0} [\frac{n^{2} + 1 + 2 n}{n^{2}}] = \frac{1}{4} lim_{n \to 0} [1 + \frac{1}{n^{n}} + \frac{2}{n}]$
$= \frac{1}{4} [1 + \frac{1}{0} + \frac{2}{0}] = \frac{1}{4}$

Karnataka CET-2019

Limits, Continuity and Differentiability

79475 The value of $lim_{x \to 0} \frac{| x |}{x}$ is

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{| x |}{x}$
LHL $= lim_{h \to 0^{-}} \frac{| 0 - h |}{0 - h} \Rightarrow lim_{h \to 0^{-}} \frac{h}{- h} = - 1$
RHL $= lim_{h \to 0^{+}} \frac{| 0 + h |}{0 + h} \Rightarrow lim_{h \to 0^{+}} \frac{h}{h} = 1$
So, LHL $\neq$ RHL
$∴$ Limit does not exist at $x = 0$

Karnataka CET-2018

Limits, Continuity and Differentiability

79476 If the function $f (x)$ satisfies $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$ , then $lim_{x \to 1} f (x) =$

1 1

2 2

3 0

4 3

Explanation:

(B) : Given, $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$
$\frac{lim_{x \to 1} (f (x) - 2)}{lim_{x \to 1} x^{2} - 1} = π$
${lim}_{x \to 1}^{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$
$lim_{x \to 1} f (x) - lim_{x \to 1} 2 = π (1 - 1)$
$lim_{x \to 1} f (x) - 2 = 0$
$lim_{x \to 1} f (x) = 2$

Karnataka CET-2014

Limits, Continuity and Differentiability

79480 $lim_{n \to \infty} {n \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} =$

1

\frac{2 π}{3}

2

\frac{π}{6}

3

\frac{π}{3}

4 1

Explanation:

(A) : Given, $lim_{n \to \infty} {n \cdot \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {n \cdot 2 \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} = \frac{1}{2} lim_{n \to \infty} {n \cdot \sin \frac{4 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {\frac{n \cdot \sin \frac{4 π}{3 n}}{\frac{4 π}{3 n}} \times \frac{4 π}{3 n}} = \frac{4 π}{6} lim_{n \to \infty} {\frac{\sin 4 \frac{π}{3 n}}{\frac{4 π}{3 n}}}$
$∵ lim_{x \to \infty} \frac{\sin x}{x} = 1]$
$= \frac{4 π}{6} \times 1 = \frac{2 π}{3}$

Limits, Continuity and Differentiability

79481 $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$

1

\frac{2}{3}

2

\frac{2}{\sqrt{3}}

3

\frac{3 \sqrt{3}}{2}

4

\frac{2}{3 \sqrt{3}}

Explanation:

(D) : Given, $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$
$lim_{x \to a} \sqrt{a + 2 x} - \sqrt{3 x} \cdot \frac{\sqrt{a + 2 x} + \sqrt{3 x}}{\sqrt{a + 2 x} + \sqrt{3 x}} \cdot \frac{1}{\sqrt{3 a + x} - 2 \sqrt{x}} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{3 a + x} + 2 \sqrt{x}}$
$= lim_{x \to a} \frac{[(a + 2 x) - 3 x]}{[(3 a + x) - 4 x]} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{(a - x)}{3 (a - x)} \times \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{1}{3} \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= \frac{1}{3} \times \frac{\sqrt{3 a + a} + 2 \sqrt{a}}{\sqrt{a + 2 a} + \sqrt{3 a}} = \frac{1}{3} \times \frac{\sqrt{4 a} + 2 \sqrt{a}}{\sqrt{3 a} + \sqrt{3 a}}$
$= \frac{1}{3} \times \frac{2 \sqrt{a} + 2 \sqrt{a}}{2 \sqrt{3 a}} = \frac{1}{3} \times \frac{4 \sqrt{a}}{2 \sqrt{3 a}} = \frac{2}{3 \sqrt{3}}$

Karnataka CET-2011

Limits, Continuity and Differentiability

79474 $\sum_{r = 1}^{n} (2 r - 1) = x$ , then
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots . . + \frac{n^{3}}{x^{2}}] =$

1

\frac{1}{2}

2 1

3

\frac{1}{4}

4 4

Explanation:

(C) : Given, $\sum_{r = 1}^{n} (2 r - 1) = x$
$2 \sum_{r = 1}^{n} r - \sum 1 = x$ s
$2 [\frac{n (n + 1)}{2}] - n$
$[\begin{matrix} 2 (n + 1) \\ x_{4} \end{matrix}] - n = x \Rightarrow n^{2} + n - n = x \Rightarrow n^{2} = x$
$x = n$
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots \dots . . + \frac{n^{3}}{x^{2}}]$
$= lim_{n \to 0} [\frac{1}{x^{2}} (1^{3} + 2^{3} + 3^{3} + \dots \dots \dots + n^{3})]$
$= lim_{n \to 0} [\frac{1}{n^{4}} \times {\frac{n (n + 1)}{2}}^{2}] = \frac{1}{4} lim_{n \to 0} [\frac{(n + 1)^{2}}{n^{2}}]$
$= \frac{1}{4} lim_{n \to 0} [\frac{n^{2} + 1 + 2 n}{n^{2}}] = \frac{1}{4} lim_{n \to 0} [1 + \frac{1}{n^{n}} + \frac{2}{n}]$
$= \frac{1}{4} [1 + \frac{1}{0} + \frac{2}{0}] = \frac{1}{4}$

Karnataka CET-2019

Limits, Continuity and Differentiability

79475 The value of $lim_{x \to 0} \frac{| x |}{x}$ is

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{| x |}{x}$
LHL $= lim_{h \to 0^{-}} \frac{| 0 - h |}{0 - h} \Rightarrow lim_{h \to 0^{-}} \frac{h}{- h} = - 1$
RHL $= lim_{h \to 0^{+}} \frac{| 0 + h |}{0 + h} \Rightarrow lim_{h \to 0^{+}} \frac{h}{h} = 1$
So, LHL $\neq$ RHL
$∴$ Limit does not exist at $x = 0$

Karnataka CET-2018

Limits, Continuity and Differentiability

79476 If the function $f (x)$ satisfies $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$ , then $lim_{x \to 1} f (x) =$

1 1

2 2

3 0

4 3

Explanation:

(B) : Given, $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$
$\frac{lim_{x \to 1} (f (x) - 2)}{lim_{x \to 1} x^{2} - 1} = π$
${lim}_{x \to 1}^{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$
$lim_{x \to 1} f (x) - lim_{x \to 1} 2 = π (1 - 1)$
$lim_{x \to 1} f (x) - 2 = 0$
$lim_{x \to 1} f (x) = 2$

Karnataka CET-2014

Limits, Continuity and Differentiability

79480 $lim_{n \to \infty} {n \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} =$

1

\frac{2 π}{3}

2

\frac{π}{6}

3

\frac{π}{3}

4 1

Explanation:

(A) : Given, $lim_{n \to \infty} {n \cdot \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {n \cdot 2 \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} = \frac{1}{2} lim_{n \to \infty} {n \cdot \sin \frac{4 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {\frac{n \cdot \sin \frac{4 π}{3 n}}{\frac{4 π}{3 n}} \times \frac{4 π}{3 n}} = \frac{4 π}{6} lim_{n \to \infty} {\frac{\sin 4 \frac{π}{3 n}}{\frac{4 π}{3 n}}}$
$∵ lim_{x \to \infty} \frac{\sin x}{x} = 1]$
$= \frac{4 π}{6} \times 1 = \frac{2 π}{3}$

Limits, Continuity and Differentiability

79481 $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$

1

\frac{2}{3}

2

\frac{2}{\sqrt{3}}

3

\frac{3 \sqrt{3}}{2}

4

\frac{2}{3 \sqrt{3}}

Explanation:

(D) : Given, $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$
$lim_{x \to a} \sqrt{a + 2 x} - \sqrt{3 x} \cdot \frac{\sqrt{a + 2 x} + \sqrt{3 x}}{\sqrt{a + 2 x} + \sqrt{3 x}} \cdot \frac{1}{\sqrt{3 a + x} - 2 \sqrt{x}} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{3 a + x} + 2 \sqrt{x}}$
$= lim_{x \to a} \frac{[(a + 2 x) - 3 x]}{[(3 a + x) - 4 x]} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{(a - x)}{3 (a - x)} \times \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{1}{3} \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= \frac{1}{3} \times \frac{\sqrt{3 a + a} + 2 \sqrt{a}}{\sqrt{a + 2 a} + \sqrt{3 a}} = \frac{1}{3} \times \frac{\sqrt{4 a} + 2 \sqrt{a}}{\sqrt{3 a} + \sqrt{3 a}}$
$= \frac{1}{3} \times \frac{2 \sqrt{a} + 2 \sqrt{a}}{2 \sqrt{3 a}} = \frac{1}{3} \times \frac{4 \sqrt{a}}{2 \sqrt{3 a}} = \frac{2}{3 \sqrt{3}}$

Karnataka CET-2011

Limits, Continuity and Differentiability

79474 $\sum_{r = 1}^{n} (2 r - 1) = x$ , then
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots . . + \frac{n^{3}}{x^{2}}] =$

1

\frac{1}{2}

2 1

3

\frac{1}{4}

4 4

Explanation:

(C) : Given, $\sum_{r = 1}^{n} (2 r - 1) = x$
$2 \sum_{r = 1}^{n} r - \sum 1 = x$ s
$2 [\frac{n (n + 1)}{2}] - n$
$[\begin{matrix} 2 (n + 1) \\ x_{4} \end{matrix}] - n = x \Rightarrow n^{2} + n - n = x \Rightarrow n^{2} = x$
$x = n$
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots \dots . . + \frac{n^{3}}{x^{2}}]$
$= lim_{n \to 0} [\frac{1}{x^{2}} (1^{3} + 2^{3} + 3^{3} + \dots \dots \dots + n^{3})]$
$= lim_{n \to 0} [\frac{1}{n^{4}} \times {\frac{n (n + 1)}{2}}^{2}] = \frac{1}{4} lim_{n \to 0} [\frac{(n + 1)^{2}}{n^{2}}]$
$= \frac{1}{4} lim_{n \to 0} [\frac{n^{2} + 1 + 2 n}{n^{2}}] = \frac{1}{4} lim_{n \to 0} [1 + \frac{1}{n^{n}} + \frac{2}{n}]$
$= \frac{1}{4} [1 + \frac{1}{0} + \frac{2}{0}] = \frac{1}{4}$

Karnataka CET-2019

Limits, Continuity and Differentiability

79475 The value of $lim_{x \to 0} \frac{| x |}{x}$ is

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{| x |}{x}$
LHL $= lim_{h \to 0^{-}} \frac{| 0 - h |}{0 - h} \Rightarrow lim_{h \to 0^{-}} \frac{h}{- h} = - 1$
RHL $= lim_{h \to 0^{+}} \frac{| 0 + h |}{0 + h} \Rightarrow lim_{h \to 0^{+}} \frac{h}{h} = 1$
So, LHL $\neq$ RHL
$∴$ Limit does not exist at $x = 0$

Karnataka CET-2018

Limits, Continuity and Differentiability

79476 If the function $f (x)$ satisfies $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$ , then $lim_{x \to 1} f (x) =$

1 1

2 2

3 0

4 3

Explanation:

(B) : Given, $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$
$\frac{lim_{x \to 1} (f (x) - 2)}{lim_{x \to 1} x^{2} - 1} = π$
${lim}_{x \to 1}^{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$
$lim_{x \to 1} f (x) - lim_{x \to 1} 2 = π (1 - 1)$
$lim_{x \to 1} f (x) - 2 = 0$
$lim_{x \to 1} f (x) = 2$

Karnataka CET-2014

Limits, Continuity and Differentiability

79480 $lim_{n \to \infty} {n \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} =$

1

\frac{2 π}{3}

2

\frac{π}{6}

3

\frac{π}{3}

4 1

Explanation:

(A) : Given, $lim_{n \to \infty} {n \cdot \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {n \cdot 2 \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} = \frac{1}{2} lim_{n \to \infty} {n \cdot \sin \frac{4 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {\frac{n \cdot \sin \frac{4 π}{3 n}}{\frac{4 π}{3 n}} \times \frac{4 π}{3 n}} = \frac{4 π}{6} lim_{n \to \infty} {\frac{\sin 4 \frac{π}{3 n}}{\frac{4 π}{3 n}}}$
$∵ lim_{x \to \infty} \frac{\sin x}{x} = 1]$
$= \frac{4 π}{6} \times 1 = \frac{2 π}{3}$

Limits, Continuity and Differentiability

79481 $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$

1

\frac{2}{3}

2

\frac{2}{\sqrt{3}}

3

\frac{3 \sqrt{3}}{2}

4

\frac{2}{3 \sqrt{3}}

Explanation:

(D) : Given, $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$
$lim_{x \to a} \sqrt{a + 2 x} - \sqrt{3 x} \cdot \frac{\sqrt{a + 2 x} + \sqrt{3 x}}{\sqrt{a + 2 x} + \sqrt{3 x}} \cdot \frac{1}{\sqrt{3 a + x} - 2 \sqrt{x}} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{3 a + x} + 2 \sqrt{x}}$
$= lim_{x \to a} \frac{[(a + 2 x) - 3 x]}{[(3 a + x) - 4 x]} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{(a - x)}{3 (a - x)} \times \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{1}{3} \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= \frac{1}{3} \times \frac{\sqrt{3 a + a} + 2 \sqrt{a}}{\sqrt{a + 2 a} + \sqrt{3 a}} = \frac{1}{3} \times \frac{\sqrt{4 a} + 2 \sqrt{a}}{\sqrt{3 a} + \sqrt{3 a}}$
$= \frac{1}{3} \times \frac{2 \sqrt{a} + 2 \sqrt{a}}{2 \sqrt{3 a}} = \frac{1}{3} \times \frac{4 \sqrt{a}}{2 \sqrt{3 a}} = \frac{2}{3 \sqrt{3}}$

Karnataka CET-2011

Limits, Continuity and Differentiability

79474 $\sum_{r = 1}^{n} (2 r - 1) = x$ , then
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots . . + \frac{n^{3}}{x^{2}}] =$

1

\frac{1}{2}

2 1

3

\frac{1}{4}

4 4

Explanation:

(C) : Given, $\sum_{r = 1}^{n} (2 r - 1) = x$
$2 \sum_{r = 1}^{n} r - \sum 1 = x$ s
$2 [\frac{n (n + 1)}{2}] - n$
$[\begin{matrix} 2 (n + 1) \\ x_{4} \end{matrix}] - n = x \Rightarrow n^{2} + n - n = x \Rightarrow n^{2} = x$
$x = n$
$lim_{n \to 0} [\frac{1^{3}}{x^{2}} + \frac{2^{3}}{x^{2}} + \frac{3^{3}}{x^{2}} + \dots \dots . . + \frac{n^{3}}{x^{2}}]$
$= lim_{n \to 0} [\frac{1}{x^{2}} (1^{3} + 2^{3} + 3^{3} + \dots \dots \dots + n^{3})]$
$= lim_{n \to 0} [\frac{1}{n^{4}} \times {\frac{n (n + 1)}{2}}^{2}] = \frac{1}{4} lim_{n \to 0} [\frac{(n + 1)^{2}}{n^{2}}]$
$= \frac{1}{4} lim_{n \to 0} [\frac{n^{2} + 1 + 2 n}{n^{2}}] = \frac{1}{4} lim_{n \to 0} [1 + \frac{1}{n^{n}} + \frac{2}{n}]$
$= \frac{1}{4} [1 + \frac{1}{0} + \frac{2}{0}] = \frac{1}{4}$

Karnataka CET-2019

Limits, Continuity and Differentiability

79475 The value of $lim_{x \to 0} \frac{| x |}{x}$ is

1 1

2 -1

3 0

4 Does not exist

Explanation:

(D) : Given, $lim_{x \to 0} \frac{| x |}{x}$
LHL $= lim_{h \to 0^{-}} \frac{| 0 - h |}{0 - h} \Rightarrow lim_{h \to 0^{-}} \frac{h}{- h} = - 1$
RHL $= lim_{h \to 0^{+}} \frac{| 0 + h |}{0 + h} \Rightarrow lim_{h \to 0^{+}} \frac{h}{h} = 1$
So, LHL $\neq$ RHL
$∴$ Limit does not exist at $x = 0$

Karnataka CET-2018

Limits, Continuity and Differentiability

79476 If the function $f (x)$ satisfies $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$ , then $lim_{x \to 1} f (x) =$

1 1

2 2

3 0

4 3

Explanation:

(B) : Given, $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$
$\frac{lim_{x \to 1} (f (x) - 2)}{lim_{x \to 1} x^{2} - 1} = π$
${lim}_{x \to 1}^{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$
$lim_{x \to 1} f (x) - lim_{x \to 1} 2 = π (1 - 1)$
$lim_{x \to 1} f (x) - 2 = 0$
$lim_{x \to 1} f (x) = 2$

Karnataka CET-2014

Limits, Continuity and Differentiability

79480 $lim_{n \to \infty} {n \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} =$

1

\frac{2 π}{3}

2

\frac{π}{6}

3

\frac{π}{3}

4 1

Explanation:

(A) : Given, $lim_{n \to \infty} {n \cdot \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {n \cdot 2 \sin \frac{2 π}{3 n} \cdot \cos \frac{2 π}{3 n}} = \frac{1}{2} lim_{n \to \infty} {n \cdot \sin \frac{4 π}{3 n}}$
$= \frac{1}{2} lim_{n \to \infty} {\frac{n \cdot \sin \frac{4 π}{3 n}}{\frac{4 π}{3 n}} \times \frac{4 π}{3 n}} = \frac{4 π}{6} lim_{n \to \infty} {\frac{\sin 4 \frac{π}{3 n}}{\frac{4 π}{3 n}}}$
$∵ lim_{x \to \infty} \frac{\sin x}{x} = 1]$
$= \frac{4 π}{6} \times 1 = \frac{2 π}{3}$

Limits, Continuity and Differentiability

79481 $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$

1

\frac{2}{3}

2

\frac{2}{\sqrt{3}}

3

\frac{3 \sqrt{3}}{2}

4

\frac{2}{3 \sqrt{3}}

Explanation:

(D) : Given, $lim_{x \to a} \frac{\sqrt{a + 2 x} - \sqrt{3 x}}{\sqrt{3 a + x} - 2 \sqrt{x}} =$
$lim_{x \to a} \sqrt{a + 2 x} - \sqrt{3 x} \cdot \frac{\sqrt{a + 2 x} + \sqrt{3 x}}{\sqrt{a + 2 x} + \sqrt{3 x}} \cdot \frac{1}{\sqrt{3 a + x} - 2 \sqrt{x}} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{3 a + x} + 2 \sqrt{x}}$
$= lim_{x \to a} \frac{[(a + 2 x) - 3 x]}{[(3 a + x) - 4 x]} \cdot \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{(a - x)}{3 (a - x)} \times \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= lim_{x \to a} \frac{1}{3} \frac{\sqrt{3 a + x} + 2 \sqrt{x}}{\sqrt{a + 2 x} + \sqrt{3 x}}$
$= \frac{1}{3} \times \frac{\sqrt{3 a + a} + 2 \sqrt{a}}{\sqrt{a + 2 a} + \sqrt{3 a}} = \frac{1}{3} \times \frac{\sqrt{4 a} + 2 \sqrt{a}}{\sqrt{3 a} + \sqrt{3 a}}$
$= \frac{1}{3} \times \frac{2 \sqrt{a} + 2 \sqrt{a}}{2 \sqrt{3 a}} = \frac{1}{3} \times \frac{4 \sqrt{a}}{2 \sqrt{3 a}} = \frac{2}{3 \sqrt{3}}$

Karnataka CET-2011

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: