11th NCERT/CBSE Limits and Derivatives Exercise 12.1 Q17to Q32
Hi

Question (17)

Evaluate the Given limit :   \[\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}\]

Solution

\[\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}\] \[ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \left( {1 - \cos 2x} \right)}}{{ - \left( {1 - \cos x} \right)}}\] \[ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}x}}{{1 - \cos x}}\] \[ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {1 - {{\cos }^2}x} \right)}}{{1 - \cos x}}\] \[ = 2\mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}{{1 - \cos x}}\] \[ = 2\mathop {\lim }\limits_{x \to 0} 1 + \cos x\] \[ = 2\left( {1 + \cos 0} \right)\] \[ = 2\left( {1 + 1} \right)\] \[ = 4\]

Question (18)

Evaluate the Given limit :   \[\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}\]

Solution

\[\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}\] Divide by x to numerator and denominator. \[\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ax}}{x} + \frac{{x\cos x}}{x}}}{{b\frac{{\sin x}}{x}}}\] \[ = \mathop {\lim }\limits_{x \to 0} \frac{{a + \cos x}}{{b\frac{{\sin x}}{x}}}\] \[ = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {a + \cos x} \right)}}{{b\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}}}\] \[ = \frac{{a + \cos 0}}{b}\] \[ = \frac{{a + 1}}{b}\]

Question (19)

Evaluate the Given limit :   \[\mathop {\lim }\limits_{x \to 0} x\sec x\]

Solution

\[\mathop {\lim }\limits_{x \to 0} x\sec x\] \[ = 0\left( {\sec 0} \right)\] \[ = 0\]

Question (20)

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

Solution

\[\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + sinbx}},\left( {a,b,a + b \ne 0} \right)\] Divide by x to numerator and denominator \[ = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin ax}}{x} + b}}{{a + \frac{{\sin bx}}{x}}}\] \[ = \frac{{\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}}} \right) \times a + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + \left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}}} \right) \times b}}\] \[ = \frac{{a + b}}{{a + b}}\] \[ = 1\]

Question (21)

Evaluate the Given limit :   \[\mathop {\lim }\limits_{x \to 0} \left( {{\mathop{\rm cosec}\nolimits} x - \cot x} \right)\]

Solution

\[\mathop {\lim }\limits_{x \to 0} \left( {{\mathop{\rm cosec}\nolimits} x - \cot x} \right)\] \[ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{\sin x}} - \frac{{\cos x}}{{\sin x}}} \right)\] \[ = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{\sin x}}\] \[ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2}}}{{2\sin \frac{x}{2}\cos \frac{x}{2}}}\] \[ = \mathop {\lim }\limits_{x \to 0} \tan \frac{x}{2}\] \[ = \tan 0\] \[ = 0\]

Question (22)

Evaluate the Given limit :   \[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}\]

Solution

\[\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}\] \[let\;x - \frac{\pi }{2} = \theta \Rightarrow x = \frac{\pi }{2} + \theta \] \[As\;x \to \frac{\pi }{2},\theta \to 0\] \[ = \mathop {\lim }\limits_{\theta \to 0} \frac{{\tan 2\left( {\frac{\pi }{2} + \theta } \right)}}{\theta }\] \[ = \mathop {\lim }\limits_{\theta \to 0} \frac{{\tan \left( {\pi + 2\theta } \right)}}{\theta }\] \[ = \mathop {\lim }\limits_{\theta \to 0} \frac{{\tan \left( {2\theta } \right)}}{\theta }\] \[ = \left[ {\mathop {\lim }\limits_{\theta \to 0} \frac{{\tan \left( {2\theta } \right)}}{{2\theta }}} \right] \times 2\] \[ = 1 \times 2\] \[ = 2\]

Question (23)

Find $\mathop {\lim }\limits_{x \to 0} f\left( x \right)and\mathop {\lim }\limits_{x \to 1} f\left( x \right)$  where
\[f(x) = \left\{ {\begin{array}{*{20}{l}}{2x + 3,}&{{\rm{x}} \le {\rm{0}}}\\{3\left( {x + 1} \right),}&{{\rm{x < 0}}}\end{array}} \right.\]

Solution

\[[f(x) = \left\{ {\begin{array}{*{20}{l}}{2x + 3,}&{{\rm{x}} \le {\rm{0}}}\\{3\left( {x + 1} \right),}&{{\rm{x}} < {\rm{0}}}\end{array}} \right.\] \[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ - }} 2x + 3\] \[ = 2\left( 0 \right) + 3\] \[ = 3\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ + }} 3\left( {x + 1} \right)\] \[ = 3\left( {0 + 1} \right)\] \[ = 3\] \[LHL = RHL\] \[\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 3\] \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} 3\left( {x + 1} \right)\] \[ = 3\left( {1 + 1} \right)\] \[ = 6\]

Question (24)

Find $\mathop {\lim }\limits_{x \to 1} f\left( x \right)$, where
\[f(x) = \left\{ {\begin{array}{*{20}{l}}{{x^2} - 1,}&{{\rm{x}} \le {\rm{1}}}\\{ - {x^2} - 1,}&{{\rm{x > 1}}}\end{array}} \right.\]

Solution

\[f(x) = \left\{ {\begin{array}{*{20}{l}}{{x^2} - 1,}&{{\rm{x}} \le {\rm{1}}}\\{ - {x^2} - 1,}&{{\rm{x > 1}}}\end{array}} \right.\] \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {1^ - }} {x^2} - 1\] \[ = {\left( 1 \right)^2} - 1\] \[ = 1 - 1 = 0\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {1^ + }} - {x^2} - 1\] \[ = - {\left( 1 \right)^2} - 1\] \[ = - 1 - 1 = - 2\] Since LHL ≠ RHL.
Limit does not exists for x =1.

Question (25)

Evaluate $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$, where
\[f(x) = \left\{ {\begin{array}{*{20}{l}}{\frac{{\left| x \right|}}{x},}&{{\rm{x}} \ne 0}\\{0,}&{{\rm{x = 0}}}\end{array}} \right.\]

Solution

\[f(x) = \left\{ {\begin{array}{*{20}{l}}{\frac{{\left| x \right|}}{x},}&{{\rm{x}} \ne 0}\\{0,}&{{\rm{x = 0}}}\end{array}} \right.\] \[\left| x \right| = x,\;\;x > 0\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\left| x \right|}}{x}\] \[ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{x}\] \[ = 1\] \[\left| x \right| = - x,\;\;x < 0\] \[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\left| x \right|}}{x}\] \[ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - x}}{x}\] \[ = - 1\] Since LHL ≠ RHL.
Limit does not exists at x = 0.

Question (26)

Find $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ where
\[f(x) = \left\{ {\begin{array}{*{20}{l}}{\frac{x}{{\left| x \right|}},}&{{\rm{x}} \ne 0}\\{0,}&{{\rm{x = 0}}}\end{array}} \right.\]

Solution

\[f(x) = \left\{ {\begin{array}{*{20}{l}}{\frac{x}{{\left| x \right|}},}&{{\rm{x}} \ne 0}\\{0,}&{{\rm{x = 0}}}\end{array}} \right.\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\left| x \right|}}\] \[ = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{x}\] \[ = 1\] \[\left| x \right| = - x,\;\;x < 0\] \[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{x}{{\left| x \right|}}\] \[ = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - x}}{x}\] \[ = - 1\] Since LHL ≠ RHL.
Limit does not exists at x = 0.

Question (27)

Find $\mathop {\lim }\limits_{x \to 5} f\left( x \right)$, where f(x) = |x| - 5

Solution

f(x) = |x| - 5
\[LHL = \mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {5^ - }} \left| x \right| - 5\] \[ = \left| 5 \right| - 5 = 0\] \[RHL = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {5^ + }} \left| x \right| - 5\] \[ = \left| 5 \right| - 5 = 0\] Since LHL = RHL
limit exists at x = 5.
\[\mathop {\lim }\limits_{x \to 5} f\left( x \right) = 0\]

Question (28)

\[\text{Suppose}\; \;f(x) = \left\{ \begin{array}{l}a + bx,\quad \,x < 1\\4,\quad \quad \quad x = 1\\b - ax\quad \;\,x > 1\end{array} \right.\] and if $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)$ what are possible values of a and b?

Solution

f(x) = \left\{ \begin{array}{l}a + bx,\quad \,x < 1\\4,\quad \quad \quad x = 1\\b - ax\quad \;\,x > 1\end{array} \right.\] and if $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)$ \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {1^ - }} a + bx\] \[ = a + b\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {1^ + }} b - ax\] \[ = b - a\] Since limit exists at x = 1, LHL = RHL.
\[a + b = b - a\] \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\] \[a + b = b - a = 4\] \[a + b = 4 - - - (1)\] \[b - a = 4 - - - (2)\] \[Add\;(1)\;and\;(2)\] \[2b = 8\] \[b = 4\] Substituting the value of b = 4 in equation (1) we get,
\[a + 4 = 4\] \[a = 0\]

Question (29)

Let a1, a2, ........,an be fixed numbers and define a function
f(x) = (x - a1) (x - a2) ..... (x - an)
What is $\mathop {\lim }\limits_{x \to {a_1}} f\left( x \right)?$ For some a ≠ a1, a2, ........,an,
compute $\mathop {\lim }\limits_{x \to a} f\left( x \right)$

Solution

\[f\left( x \right) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right)\left( {x - {a_3}} \right).........\left( {x - {a_n}} \right)\] \[\mathop {\lim }\limits_{x \to {a_1}} f\left( x \right) = \mathop {\lim }\limits_{x \to {a_1}} \left( {x - {a_1}} \right)\left( {x - {a_2}} \right)\left( {x - {a_3}} \right).........\left( {x - {a_n}} \right)\] \[ = \left( {{a_1} - {a_1}} \right)\left( {{a_1} - {a_2}} \right)\left( {{a_1} - {a_3}} \right).........\left( {{a_1} - {a_n}} \right)\] \[ = 0\] \[\mathop {\lim }\limits_{x \to a} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to a} \left( {x - {a_1}} \right)\left( {x - {a_2}} \right)\left( {x - {a_3}} \right).........\left( {x - {a_n}} \right)\] \[ = \left( {a - {a_1}} \right)\left( {a - {a_2}} \right)\left( {a - {a_3}} \right).........\left( {a - {a_n}} \right)\]

Question (30)

\[ \text{If}\;f(x) = \left\{ \begin{array}{l}\left| x \right| + 1,\quad \,x < 0\\0,\quad \quad \quad x = 0\\\left| x \right| - 1,\quad \;\,x > 0\end{array} \right.\] For what value (s) of a does $\mathop {\lim }\limits_{x \to a} f\left( x \right)\]

Solution

\;f(x) = \left\{ \begin{array}{l}\left| x \right| + 1,\quad \,x < 0\\0,\quad \quad \quad x = 0\\\left| x \right| - 1,\quad \;\,x > 0\end{array} \right.\] \[at\;x = 0\] \[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ - }} \left| x \right| + 1\] \[ = 0 + 1 = 1\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ + }} \left| x \right| - 1\] \[ = 0 - 1 = - 1\] Since LHL ≠ RHL
limit does not exists for x = 0.
So a ≠ 0
For a < 0,
| x | = -x.
\[LHL = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} \left| x \right| + 1\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} - x + 1\] \[ = - a + 1\] \[RHL = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} \left| x \right| + 1\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} - x + 1\] \[ = - a + 1\] Since LHL = RHL ,
So limit exists for all a < 0
For a > 0
|x| = x \[LHL = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} \left| x \right| + 1\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} x + 1\] \[ = a + 1\] \[RHL = \mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} \left| x \right| + 1\] \[ = \mathop {\lim }\limits_{x \to {a^ - }} x + 1\] \[ = a + 1\] Since LHL = RHL ,
So limit exists for all a > 0.
Limit will exists for the values of a other than 0.

Question (31)

If the function f(x) satisfies $\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{{{x^2} - 1}} = \pi $, evalulate $\mathop {\lim }\limits_{x \to 1} f\left( x \right)$

Solution

\[\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{{{x^2} - 1}} = \pi \] \[\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right) - 2}}{{\mathop {\lim }\limits_{x \to 1} {x^2} - 1}} = \pi \] \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) - 2 = \pi \left[ {\mathop {\lim }\limits_{x \to 1} {x^2} - 1} \right]\] \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) - 2 = \pi \left[ {{1^2} - 1} \right]\] \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) - 2 = 0\] \[\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 2\]

Question (32)

\[{\rm{If,}}f(x) = \left\{ {\begin{array}{*{20}{l}}{m{x^2} + n,\quad {\mkern 1mu} \;x < 0}\\{nx + m,\quad \quad 0 \le x \le 1}\\{n{x^3} + m,\quad \;{\mkern 1mu} \;\;\;x > 1}\end{array}} \right.\]For what integers m and n does both $\mathop {\lim }\limits_{x \to 0} f\left( x \right)\;and\;\mathop {\lim }\limits_{x \to 1} f\left( x \right)$, exists?

Solution

\[{\rm{If,}}f(x) = \left\{ {\begin{array}{*{20}{l}}{m{x^2} + n,\quad {\mkern 1mu} \;x < 0}\\{nx + m,\quad \quad 0 \le x \le 1}\\{n{x^3} + m,\quad \;{\mkern 1mu} \;\;\;x > 1}\end{array}} \right.\] \[For\;x = 0\] \[LHL = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ - }} m{x^2} + n\] \[ = 0 + n = n\] \[RHL = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {0^ + }} nx + m\] \[ = 0 + m = m\] Since limit exists for f(x) at x = 0,
LHL = RHL
So m = n. ---(1)
\[For\;x = 1\] \[LHL = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {1^ - }} nx + m\] \[ = n\left( 1 \right) + m\] \[ = m + n\] \[RHL = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\] \[ = \mathop {\lim }\limits_{x \to {1^ + }} n{x^3} + m\] \[ = n\left( 1 \right) + m\] \[ = m + n\] Since limit exists at x = 1,
LHL = RHL
So m + n = m + n
Which is true for all values of m and n.
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⇒13.1(Q17 t0 Q32)