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