11th NCERT/CBSE Limits and Derivatives Exercise 12.1 Q1 to Q16
Hi

Question (1)

Evaluate the Given limit : $\mathop {\lim }\limits_{x \to 3} \left( {x + 3} \right)$

Solution

$\mathop {\lim }\limits_{x \to 3} \left( {x + 3} \right) = 3 + 3 = 6$

Question (2)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to \pi } \left( {x - \frac{{22}}{7}} \right)$

Solution

$\mathop {\lim }\limits_{x \to \pi } \left( {x - \frac{{22}}{7}} \right) = \left( {\pi - \frac{{22}}{7}} \right)$

Question (3)

Evaluate the Given limit :   $\mathop {\lim }\limits_{r \to 1} \left( {\pi {r^2}} \right)$

Solution

$\mathop {\lim }\limits_{r \to 1} \left( {\pi {r^2}} \right) = \pi {\left( 1 \right)^2} = \pi$

Question (4)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 4} \left( {\frac{{4x + 3}}{{x - 2}}} \right)$

Solution

$\mathop {\lim }\limits_{x \to 4} \left( {\frac{{4x + 3}}{{x - 2}}} \right) = \frac{{4\left( 4 \right) + 3}}{{4 - 2}} = \frac{{19}}{2}$

Question (5)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to - 1} \left( {\frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}} \right)$

Solution

$\mathop {\lim }\limits_{x \to - 1} \left( {\frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}} \right) = \frac{{{{\left( { - 1} \right)}^{10}} + {{\left( { - 1} \right)}^5} + 1}}{{ - 1 - 1}} = - \frac{1}{2}$

Question (6)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x}$

Solution

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x}$ Put x + 1 = y so that Y → 1 as x → 0
Accordiingly,  $\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x} = \mathop {\lim }\limits_{y \to 1} \frac{{{{\left( y \right)}^5} - 1}}{{y - 1}} = \mathop {\lim }\limits_{y \to 1} \frac{{{{\left( y \right)}^5} - {{\left( 1 \right)}^5}}}{{y - 1}}$
Use following formula $\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$
$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x} = 5 \cdot {1^{5 - 1}} = 5$ $\therefor \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x} = 5$

Question (7)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}}$

Solution

At x = 2, the value of the given rational function takes the form $\frac{0}{0}$
$\therefore \mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {3x + 5} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}$ $= \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3x + 5} \right)}}{{\left( {x + 2} \right)}}$ $= \frac{{3\left( 2 \right) + 5}}{{2 + 2}} = \frac{{11}}{4}$

Question (8)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}}$

Solution

At x = 2, the value of the given rational function takes the form $\frac{0}{0}$
$\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x + 3} \right)\left( {{x^2} + 9} \right)}}{{\left( {x + 3} \right)\left( {2x + 1} \right)}}$ $= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x + 3} \right)\left( {{x^2} + 9} \right)}}{{2x + 1}}$ $= \frac{{\left( {3 + 3} \right)\left( {{3^2} + 9} \right)}}{{2\left( 3 \right) + 1}}$ $= \frac{{6\left( {18} \right)}}{7} = \frac{{108}}{7}$

Question (9)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}}$

Solution

$\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}} = \frac{{a\left( 0 \right) + b}}{{c\left( 0 \right) + 1}} = b$

Question (10)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 1} \left( {\frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}} \right)$

Solution

$\mathop {\lim }\limits_{x \to 1} \left( {\frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}} \right)$ At z = 1, the value of the given function takes the form $\frac{0}{0}$.
Put ${z^{\frac{1}{6}}} = x$ so that z →1 as x → 1.
Accordingly,
$\mathop {\lim }\limits_{x \to 1} \left( {\frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}}$ $= \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - {1^2}}}{{x - 1}}$
Use following formula $\mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$
$= 2 \cdot {1^{2 - 1}} = 2$ $\therefore \mathop {\lim }\limits_{x \to 1} \left( {\frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}} \right) = 2$

Question (11)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 1} \left( {\frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}}} \right),a + b + c \ne 0$

Solution

$\mathop {\lim }\limits_{x \to 1} \left( {\frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}}} \right) = \frac{{a{{\left( 1 \right)}^2} + b\left( 1 \right) + c}}{{c{{\left( 1 \right)}^2} + b\left( 1 \right) + a}}$ $\left[ {As\;a + b + c \ne 0} \right]$ $= \frac{{a + b + c}}{{a + b + c}} = 1$

Question (12)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to - 2} \left( {\frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}} \right)$

Solution

$\mathop {\lim }\limits_{x \to - 2} \left( {\frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}} \right)$ At x = –2, the value of the given function takes the form $\frac{0}{0}$.
$\text{Now,}\mathop {\lim }\limits_{x \to - 2} \left( {\frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}} \right) = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {\frac{{2 + x}}{{2x}}} \right)}}{{x + 2}}$ $= \mathop {\lim }\limits_{x \to - 2} \frac{1}{{2x}} = \frac{1}{{2\left( { - 2} \right)}} = \frac{{ - 1}}{4}$

Question (13)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{bx}}} \right)$

Solution

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{bx}}} \right)$ At x = 0, the value of the given function takes the form $\frac{0}{0}$
$\text{Now,}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{bx}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}} \times \frac{{ax}}{{bx}}$ $= \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right) \times \left( {\frac{a}{b}} \right)$ $\text{Note,}\left[ {x \to 0 \Rightarrow ax \to 0} \right]$ $= \frac{a}{b}\mathop {\lim }\limits_{ax \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right)$
$\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y} = 1$
$= \frac{a}{b} \times 1 = \frac{a}{b}$

Question (14)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{sin\,ax}}{{\sin bx}}} \right),a,b \ne 0$

Solution

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{sin{\mkern 1mu} ax}}{{\sin bx}}} \right),a,b \ne 0$ At x = 0, the value of the given function takes the form $\frac{0}{0}$
$\text{Now,}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{sin\,{\mkern 1mu} ax}}{{\sin bx}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\sin ax}}{{ax}}} \right) \times ax}}{{\left( {\frac{{\sin bx}}{{bx}}} \right) \times bx}}$
$x \to 0 \Rightarrow ax \to 0$ and $x \to 0 \Rightarrow bx \to 0$
$= \left( {\frac{a}{b}} \right)\frac{{\mathop {\lim }\limits_{ax \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right)}}{{\mathop {\lim }\limits_{bx \to 0} \left( {\frac{{\sin bx}}{{bx}}} \right)}}$
$\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y} = 1$
$= \left( {\frac{a}{b}} \right) \times \frac{1}{1} = \frac{a}{b}$

Question (15)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin \left( {\pi - x} \right)}}{{\pi \left( {\pi - x} \right)}}$

Solution

$\mathop {\lim }\limits_{x \to \pi } \frac{{\sin \left( {\pi - x} \right)}}{{\pi \left( {\pi - x} \right)}}$
It is seen that x → π ⇒ (π – x) → 0
$\mathop {\lim }\limits_{x \to \pi } \frac{{\sin \left( {\pi - x} \right)}}{{\pi \left( {\pi - x} \right)}} = \frac{1}{\pi }\mathop {\lim }\limits_{\left( {\pi - x} \right) \to 0} \frac{{\sin \left( {\pi - x} \right)}}{{\left( {\pi - x} \right)}}$
$\mathop {\lim }\limits_{y \to 0} \frac{{\sin y}}{y} = 1$
$= \frac{1}{\pi } \times 1 = \frac{1}{\pi }$

Question (16)

Evaluate the Given limit :   $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}}$

Solution

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}} = \frac{{\cos 0}}{{\pi - 0}} = \frac{1}{\pi }$