Chapter 13: Limits and Derivatives

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + {x^3}}  - \sqrt {1 - {x^3}} }}{{{x^2}}}$

Then, 

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + {x^3}}  - \sqrt {1 - {x^3}} }}{{{x^2}}}{\text{ }}\left( {{\text{given}}} \right)$

Rationalise with $\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + {x^3}}  - \sqrt {1 - {x^3}} }}{{{x^2}}} \cdot \dfrac{{\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} }}{{\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} }}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 + {x^3}} \right) - \left( {1 - {x^3}} \right)}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + {x^3} - 1 + {x^3}}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{x^3}}}{{{x^2}\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2x}}{{\left( {\sqrt {1 + {x^3}}  + \sqrt {1 - {x^3}} } \right)}}$

$= 0$

Ans: Given:

\[\mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^3} + 27}}{{{x^5} + 243}}\]

Then, 

\[ = \mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^3} + 27}}{{{x^5} + 243}}{\text{ }}\left( {{\text{given}}} \right)\]

\[ = \mathop {\lim }\limits_{x \to  - 3} \dfrac{{\dfrac{{{x^3} + 27}}{{x + 3}}}}{{\dfrac{{{x^5} + 243}}{{x + 3}}}}\]

\[ = \mathop {\lim }\limits_{x \to  - 3} \dfrac{{\dfrac{{{x^3} - {{( - 3)}^3}}}{{x - ( - 3)}}}}{{\dfrac{{{x^5} - {{( - 3)}^5}}}{{x - ( - 3)}}}}\]

\[ = \dfrac{{\mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^3} - {{( - 3)}^3}}}{{x - ( - 3)}}}}{{\mathop {\lim }\limits_{x \to  - 3} \dfrac{{{x^5} - {{( - 3)}^5}}}{{x - ( - 3)}}}}{\text{ }}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}} \right]\]

\[ = \dfrac{{3{{( - 3)}^{3 - 1}}}}{{5{{( - 3)}^{5 - 1}}}}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ = \dfrac{3}{5}\dfrac{{{{( - 3)}^2}}}{{{{( - 3)}^4}}}\]

\[ = \dfrac{3}{{5{{( - 3)}^2}}}\]

\[ = \dfrac{3}{{45}}\]

\[ = \dfrac{1}{{15}}\]

Ans: Given:

$\mathop {\lim }\limits_{x \to 1/2} \left( {\dfrac{{8x - 3}}{{2x - 1}} - \dfrac{{4{x^2} + 1}}{{4{x^2} - 1}}} \right)$

Then,

$= \mathop {\lim }\limits_{x \to 1/2} \left( {\dfrac{{8x - 3}}{{2x - 1}} - \dfrac{{4{x^2} + 1}}{{4{x^2} - 1}}} \right)\,\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 1/2} \left[ {\dfrac{{(8x - 3)(2x + 1) - \left( {4{x^2} + 1} \right)}}{{\left( {4{x^2} - 1} \right)}}} \right]$

$= \mathop {\lim }\limits_{x \to 1/2} \left[ {\dfrac{{16{x^2} + 8x - 6x - 3 - 4{x^2} - 1}}{{4{x^2} - 1}}} \right]$

$= \mathop {\lim }\limits_{x \to 1/2} \left[ {\dfrac{{12{x^2} + 2x - 4}}{{4{x^2} - 1}}} \right]$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2\left( {6{x^2} + x - 2} \right)}}{{4{x^2} - 1}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2\left( {6{x^2} + 4x - 3x - 2} \right)}}{{4{x^2} - 1}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2[2x(3x + 2) - 1(3x + 2)]}}{{4{x^2} - 1}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2[(3x + 2)(2x - 1)]}}{{{{(2x)}^2} - {{(1)}^2}}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2(3x + 2)(2x - 1)}}{{(2x - 1)(2x + 1)}}$

$= \mathop {\lim }\limits_{x \to 1/2} \dfrac{{2(3x + 2)}}{{2x - 1}}$

$= \dfrac{{2\left( {3 \times \dfrac{1}{2} + 2} \right)}}{{2 \times \dfrac{1}{2} + 1}}$

$= \dfrac{3}{2} + 2$

$= \dfrac{7}{2}$ 

Ans: Given:

\[\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^n} - {2^n}}}{{x - 2}} = 80,{\text{ }}n \in N\]

Then, 

\[ \Rightarrow {\text{ }}n{(2)^{n - 1}} = 80{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ \Rightarrow {\text{ }}n \times {2^{n - 1}} = 5 \times 16\]

\[ \Rightarrow {\text{ }}n \times {2^{n - 1}} = 5 \times {(2)^4}\]

\[ \Rightarrow {\text{ }}n \times {2^{n - 1}} = 5 \times {(2)^{5 - 1}}\]

\[\therefore n = 5\]

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{\sin 7x}}$

We know that,

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$\because x \to 0 \Rightarrow \left( {kx \to 0} \right),{\text{ here }}k{\text{ is real number}}$

\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{\sin 7x}}{\text{ }}\left( {{\text{given}}} \right)\]

Multiply and divide 3x in numerator and denominator 

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin 3x}}{{3x}} \cdot 3x}}{{\dfrac{{\sin 7x}}{{7x}} \cdot 7x}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 7x}}{{7x}}}} \cdot \dfrac{{3x}}{{7x}}$

$= \dfrac{3}{7} \cdot \dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{{3x}}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 7x}}{{7x}}}}{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= \dfrac{3}{7}{\text{                    }}\left[ {\because x \to 0 \Rightarrow \left( {kx \to 0} \right),{\text{ here }}k{\text{ is real number}}} \right]$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{{{\sin }^2}4x}}$

We know that, \[\because [\sin 2\theta  = 2\sin \theta \cos \theta ]\]

\[\because \cos 0 = 1\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x\mid }}{{{{[\sin 2(2x)]}^2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{{{(2\sin 2x\cos 2x)}^2}}}$

\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{4{{\sin }^2}2x{{\cos }^2}2x}}{\text{ }}\left[ {\because \sin 2\theta  = 2\sin \theta \cos \theta } \right]\]

\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{4{{\cos }^2}2x}}\left[ {\because \cos 0 = 1} \right]\]

\[ = \dfrac{1}{4}\]

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}}$

We know that, $\cos 2x = 1 - 2{\sin ^2}x$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - 1 + 2{{\sin }^2}x}}{{{x^2}}}{\text{ }}\left[ {\because \cos 2x = 1 - 2{{\sin }^2}x} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}x}}{{{x^2}}}$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}}}$ 

$= 2\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin x}}{x}} \right)^2}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 2 \times 1 = 2$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x - \sin 2x}}{{{x^3}}}$

We know that,

$\because \sin 2x = 2\sin x\cos x$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x - \sin 2x}}{{{x^3}}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x - 2\sin x\cos x}}{{{x^3}}}{\text{ }}\left[ {\because \sin 2x = 2\sin x\cos x} \right]$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{2\sin x(1 - \cos x)}}{{{x^3}}}$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{1 - \cos x}}{{{x^2}}}} \right)$

$= 2 \cdot 1\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{{x^2}}}{\text{                }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - 1 + 2{{\sin }^2}\dfrac{x}{2}}}{{{x^2}}}$

$= 2\mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{4 \times \dfrac{{{x^2}}}{4}}}$ 

$= \dfrac{{2 \cdot 2}}{4}\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2}$

$= \mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2} = 1$ 

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos mx}}{{1 - \cos nx}}$

We know that,

$\because \cos mx = 1 - 2{\sin ^2}\dfrac{{mx}}{2}{\text{and }}\sin nx = 1 - 2{\sin ^2}\dfrac{{nx}}{2}$

\[\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\]

Then,

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos mx}}{{1 - \cos nx}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - 1 + 2{{\sin }^2}\dfrac{{mx}}{2}}}{{1 - 1 + 2{{\sin }^2}\dfrac{{nx}}{2}}}$

${\because \cos mx = 1 - 2{{\sin }^2}\dfrac{{mx}}{2}}$ 

${{\text{ and }}\sin nx = 1 - 2{{\sin }^2}\dfrac{{nx}}{2}}$

$\mathop { = \lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}\dfrac{{mx}}{2}}}{{{{\sin }^2}\dfrac{{nx}}{2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{\sin }^2}\dfrac{{mx}}{2}}}{{{{\left( {\dfrac{{mx}}{2}} \right)}^2}}} \cdot {{\left( {\dfrac{{mx}}{2}} \right)}^2}}}{{\dfrac{{{{\sin }^2}\dfrac{{nx}}{2}}}{{{{\left( {\dfrac{{nx}}{2}} \right)}^2}}} \cdot {{\left( {\dfrac{{nx}}{2}} \right)}^2}}}$

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{mx}}{2}}}{{\dfrac{{mx}}{2}}}} \right)}^2}}}{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{nx}}{2}}}{{\dfrac{{nx}}{2}}}} \right)}^2}}} \cdot \dfrac{{{m^2}\dfrac{{{x^2}}}{4}}}{{{n^2}\dfrac{{{x^2}}}{4}}}$

\[ = \dfrac{{{m^2}}}{{{n^2}}} \cdot \dfrac{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{mx}}{2}}}{{\dfrac{{mx}}{2}}}} \right)}^2}}}{{\mathop {\lim }\limits_{x \to 0} {{\left( {\dfrac{{\sin \dfrac{{nx}}{2}}}{{\dfrac{{nx}}{2}}}} \right)}^2}}}{\text{  }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]\]

\[ = \dfrac{{{m^2}}}{{{n^2}}}{\text{                          }}\left[ {\because x \to 0 \Rightarrow kx \to 0} \right]\]

Ans: Given:

$\mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt {1 - \cos 6x} }}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}$

We know that,

$\because \cos 2x = 1 - 2{\sin ^2}x$

$\because \sin \left( {\pi  - \theta } \right) = \sin \theta$

$\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$

$\because x \to \dfrac{\pi }{3} \Rightarrow \left( {x - \dfrac{\pi }{3}} \right) \to 0$

Then,

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt {1 - \cos 6x} }}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}{\text{ }}\left( {{\text{given}}} \right)$

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt {1 - 1 + 2{{\sin }^2}3x} }}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}{\text{ }}\left[ {\because \cos 2x = 1 - 2{{\sin }^2}x} \right]$

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sqrt 2 \sin 3x}}{{\sqrt 2 \left( {\dfrac{\pi }{3} - x} \right)}}$

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sin 3x}}{{\dfrac{\pi }{3} - x}}$ 

$= \mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sin (\pi  - 3x)}}{{\dfrac{{\pi  - 3x}}{3}}}{\text{         }}\left[ {\because \sin \left( {\pi  - \theta } \right) = \sin \theta } \right]$

$= 3\mathop {\lim }\limits_{x \to \pi /3} \dfrac{{\sin (\pi  - 3x)}}{{(\pi  - 3x)}}{\text{       }}\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$

$= 3 \times 1$ 

$= 3{\text{  }}\left[ {\because x \to \dfrac{\pi }{3} \Rightarrow \left( {x - \dfrac{\pi }{3}} \right) \to 0} \right]$