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

Contact US     Exercise:-     13.1 Q1 to Q16     13.1 Q17 to Q32     13.2     Miscellaneous     maths index

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,  $\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x} = \mathop {\lim }\limits_{y \to 1} \frac{{{{\left( y \right)}^5} - 1}}{{y - 1}} = \mathop {\lim }\limits_{y \to 1} \frac{{{{\left( y \right)}^5} - {{\left( 1 \right)}^5}}}{{y - 1}}$
\[\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x} = 5 \cdot {1^{5 - 1}} = 5\] \[\therefor \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {x + 1} \right)}^5} - 1}}{x} = 5\]

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,
\[\mathop {\lim }\limits_{x \to 1} \left( {\frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}}\] \[ = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - {1^2}}}{{x - 1}}\] \[ = 2 \cdot {1^{2 - 1}} = 2\] \[\therefore \mathop {\lim }\limits_{x \to 1} \left( {\frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}} \right) = 2\]

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 $\frac{0}{0}$
\[\text{Now,}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{bx}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}} \times \frac{{ax}}{{bx}}\] \[ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right) \times \left( {\frac{a}{b}} \right)\] \[\text{Note,}\left[ {x \to 0 \Rightarrow ax \to 0} \right]\] \[ = \frac{a}{b}\mathop {\lim }\limits_{ax \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right)\] \[ = \frac{a}{b} \times 1 = \frac{a}{b}\]

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 $\frac{0}{0}$
\[\text{Now,}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{sin\,{\mkern 1mu} ax}}{{\sin bx}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\sin ax}}{{ax}}} \right) \times ax}}{{\left( {\frac{{\sin bx}}{{bx}}} \right) \times bx}}\] \[ = \left( {\frac{a}{b}} \right)\frac{{\mathop {\lim }\limits_{ax \to 0} \left( {\frac{{\sin ax}}{{ax}}} \right)}}{{\mathop {\lim }\limits_{bx \to 0} \left( {\frac{{\sin bx}}{{bx}}} \right)}}\] \[ = \left( {\frac{a}{b}} \right) \times \frac{1}{1} = \frac{a}{b}\]

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
\[\mathop {\lim }\limits_{x \to \pi } \frac{{\sin \left( {\pi - x} \right)}}{{\pi \left( {\pi - x} \right)}} = \frac{1}{\pi }\mathop {\lim }\limits_{\left( {\pi - x} \right) \to 0} \frac{{\sin \left( {\pi - x} \right)}}{{\left( {\pi - x} \right)}}\] \[ = \frac{1}{\pi } \times 1 = \frac{1}{\pi }\]

Question (16)

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

Solution

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