Chapter 13: Limits and Derivatives

Ans: Given:

$\mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}}$

\[\because [{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)]\]

$= \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}}{\text{ }}\left( {{\text{given}}} \right)$

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

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

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

$  = 3 + 3 = 6$ 

Ans: Given:

$\mathop {\lim }\limits_{x \to 1/2} \dfrac{{4{x^2} - 1}}{{2x - 1}}$

Then,

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

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

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

$= 2 \times \dfrac{1}{2} + 1$

$= 1 + 1$

$= 2$ 

Ans: Given:

$\mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h}  - \sqrt x }}{h}$

\[\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}\]

\[\because h \to 0 \Rightarrow x + h \to x\]

\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {x + h}  - \sqrt x }}{h}{\text{ }}\left( {{\text{given}}} \right)\]

\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^{1/2}} - {{(x)}^{1/2}}}}{{x + h - x}}\]

\[ = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^{1/2}} - {{(x)}^{1/2}}}}{{(x + h) - x}}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

\[ = \dfrac{1}{2}{x^{\dfrac{1}{2} - 1}}\]

\[ = \dfrac{1}{2}{x^{ - 1/2}}\quad {\text{  }}\left[ {\because h \to 0 \Rightarrow x + h \to x} \right]\]

\[ = \dfrac{1}{{2\sqrt x }}\]

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(x + 2)}^{1/3}} - {2^{1/3}}}}{x}{\text{ }}$

Then,

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

$= \dfrac{1}{3} \times {2^{\dfrac{1}{3} - 1}} = \dfrac{1}{3} \times {(2)^{ - 2/3}}\quad \left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$

$= \dfrac{1}{{3{{(2)}^{2/3}}}}\quad {\text{                    }}[\because x \to 0 \Rightarrow x + 2 \to 2]$

Ans: Given:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^6} - 1}}{{{{(1 + x)}^2} - 1}}$

Divide numerator and denominator by \[x\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{(1 + x)}^6} - 1}}{x}}}{{\dfrac{{{{(1 + x)}^2} - 1}}{x}}}$\[\left[ {\because x \to 0 \Rightarrow 1 + x \to 1} \right]\]

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{(1 + x)}^6} - 1}}{{(1 + x) - 1}}}}{{\dfrac{{{{(1 + x)}^2} - 1}}{{(1 + x) - 1}}}}$

$= \dfrac{{\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^6} - {{(1)}^6}}}{{(1 + x) - 1}}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 + x)}^2} - {{(1)}^2}}}{{(1 + x) - 1}}}}$$\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]$

$= \dfrac{{6{{(1)}^{6 - 1}}}}{{2{{(1)}^{2 - 1}}}}$\[\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]\]

$= \dfrac{{6 \times 1}}{{2 \times 1}}$

$= \dfrac{6}{2}$

$= 3$

Ans: Given:

$\mathop {\lim }\limits_{x \to a} \dfrac{{{{(2 + x)}^{5/2}} - {{(a + 2)}^{5/2}}}}{{x - a}}$

Then,

$= \mathop {\lim }\limits_{x \to a} \dfrac{{{{(2 + x)}^{5/2}} - {{(a + 2)}^{5/2}}}}{{(2 + x) - (a + 2)}}$

$= \dfrac{5}{2}{\left( {a + 2} \right)^{\dfrac{5}{2} - 1}}\left[ {\because \mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$

$= \dfrac{5}{2}{\left( {a + 2} \right)^{\dfrac{3}{2}}}{\text{                }}\left[ {\because x \to a \Rightarrow x + 2 \to a + 2} \right]$

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{x - \sqrt x }}{{\sqrt x  - 1}}$

Then,

Then,

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt x \left[ {{{\left( x \right)}^{\dfrac{7}{2}}} - 1} \right]}}{{\sqrt x  - 1}}$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{{{(x)}^{7/2}} - 1}}{{\sqrt x  - 1}} \cdot \mathop {\lim }\limits_{x \to 1} \sqrt x$

$\left[ {\because \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot g\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right]$

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^{7/2}} - 1}}{{x - 1}}}}{{\dfrac{{{{(x)}^{1/2}} - 1}}{{x - 1}}}} \cdot 1$

$= \dfrac{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{7/2}} - 1}}{{x - 1}}}}{{\mathop {\lim }\limits_{x \to 1} \dfrac{{{{(x)}^{1/2}} - 1}}{{x - 1}}}}$$\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]$

$= \dfrac{{\dfrac{7}{2}{{(1)}^{\dfrac{7}{2} - 1}}}}{{\dfrac{1}{2}{{(1)}^{\dfrac{1}{2} - 1}}}}$

$= \dfrac{{\dfrac{7}{2}}}{{\dfrac{1}{2}}}$

$= 7$

Ans: Given:

$\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 4}}{{\sqrt {3x - 2}  - \sqrt {x + 2} }}$

Then, 

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

$= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - 4} \right)(\sqrt {3x - 2}  + \sqrt {x + 2} )}}{{{{(\sqrt {3x - 2} )}^2} - {{(\sqrt {x + 2} )}^2}}}{\text{  }}\left[ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$

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

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

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

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

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

$= \dfrac{{(2 + 2)(\sqrt {6 - 2}  + \sqrt {2 + 2} )}}{2}$

$= \dfrac{{4(2 + 2)}}{2} = 8$

Ans: Given:     

$\mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{{x^4} - 4}}{{{x^2} + 3\sqrt 2 x - 8}}$

Then, 

$\mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{{x^4} - 4}}{{{x^2} + 3\sqrt 2 x - 8}}$

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

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

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

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

$= \mathop {\lim }\limits_{x \to \sqrt 2 } \dfrac{{(x + \sqrt 2 )\left( {{x^2} + 2} \right)}}{{(x + 4\sqrt 2 )}}$

\[ = \dfrac{{\left( {\sqrt 2  + \sqrt 2 } \right)\left[ {{{\left( {\sqrt 2 } \right)}^2} + 2} \right]}}{{\left( {\sqrt 2  + 4\sqrt 2 } \right)}}\]

\[ = \dfrac{{2\sqrt 2 \left( {2 + 2} \right)}}{{5\sqrt 2 }}\]

\[ = \dfrac{8}{5}\]

Ans: Given:

$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^7} - 2{x^5} + 1}}{{{x^3} - 3{x^2} + 2}}$

Then,

$= \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^7} - {x^5} - {x^5} + 1}}{{{x^3} - {x^2} - 2{x^2} + 2}}$

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

On dividing numerator and denominator by $\left( {x - 1} \right){\text{, then}}$

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

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

$= \dfrac{{1 \times 2 - 5 \times {{(1)}^4}}}{{1 - 2 \times 2}}$

$= \dfrac{{2 - 5}}{{1 - 4}}$

$= \dfrac{{ - 3}}{{ - 3}} = 1$