Skip to content
Related Articles

Related Articles

Improve Article
Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.5 | Set 1
  • Last Updated : 03 Mar, 2021

Evaluate the following limits:

Question 1. \lim_{x \to a} \frac {(x+2)^{\frac 5 2}-(a+2)^{\frac 5 2}} {x-a}

Solution:

\lim_{x \to a} \frac {(x+2)^{\frac 5 2}-(a+2)^{\frac 5 2}} {x-a}

=\lim_{x \to a} \frac {(x+2)^{\frac 5 2}-(a+2)^{\frac 5 2}} {(x+2)-(a+2)}

Let y = x + 2 and b = a + 2

=\lim_{y \to b} \frac {(y)^{\frac 5 2}-(b)^{\frac 5 2}} {(y)-(b)} 



=\frac 5 2 b^{{\frac 5 2} -1}    [using formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}]

=\frac 5 2 (a+2)^{{\frac 5 2} -1}

=\frac 5 2 (a+2)^{\frac 3 2}

Question 2. \lim_{x \to a} \frac {(x+2)^{\frac 3 2}-(a+2)^{\frac 3 2}} {x-a}

Solution:

\lim_{x \to a} \frac {(x+2)^{\frac 3 2}-(a+2)^{\frac 3 2}} {x-a}

=\lim_{x \to a} \frac {(x+2)^{\frac 3 2}-(a+2)^{\frac 3 2}} {(x+2)-(a+2)}

Let y=x+2 and b=a+2

=\lim_{y \to b} \frac {(y)^{\frac 3 2}-(b)^{\frac 3 2}} {(y)-(b)}



=\frac 3 2 b^{{\frac 3 2} -1}    [using formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}]

=\frac 3 2 (a+2)^{{\frac 3 2} -1}

=\frac 3 2 (a+2)^{\frac 1 2}

Question 3. \lim_{x \to 0} \frac {(1+x)^{6}-1} {(1+x)^2-1}

Solution:

\lim_{x \to 0} \frac {(1+x)^{6}-1} {(1+x)^2-1}

Dividing the numerator and denominator with 1+x-1

=\lim_{x \to 0} \frac {\frac {(1+x)^{6}-1^6} {1+x-1} } {\frac {(1+x)^{2}-1^2} {1+x-1}}

Let y=1+x as x \to 0,y\to 1

=\lim_{y \to 1} \frac {\frac {y^{6}-1^6} {y-1} } {\frac {y^{2}-1^2} {y-1}}

=\frac {6(1)^{6-1}} {2(1)^{2-1}}    [using formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}]



=\frac 6 2

=3

Question 4. \lim_{x \to a} \frac {x^{\frac 2 7}-a^{\frac 2 7}} {x-a}

Solution:

\lim_{x \to a} \frac {x^{\frac 2 7}-a^{\frac 2 7}} {x-a}

Applying the formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}, here, n=\frac 2 7

=\frac 2 7a^{{\frac 2 7}-1}

=\frac 2 7 a^{\frac {-5} {7}}

Question 5. \lim_{x \to a} \frac {x^{\frac 5 7}-a^{\frac 5 7}} {x^{\frac 2 7}-a^{\frac 2 7}}

Solution:

\lim_{x \to a} \frac {x^{\frac 5 7}-a^{\frac 5 7}} {x^{\frac 2 7}-a^{\frac 2 7}}

Dividing the numerator and denominator with x-a



=\lim_{x \to a} \frac { \frac {x^{\frac 5 7}-a^{\frac 5 7}} {x-a}} {\frac {x^{\frac 2 7}-a^{\frac 2 7}} {x-a}}

Applying the formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}, here, n=\frac 5 7   in numerator and n=\frac 2 7   in denominator 

=\frac {\frac 5 7a^{{\frac 5 7}-1}} {\frac 2 7a^{{\frac 2 7}-1}}

=\frac {\frac 5 7a^{\frac {-2} 7}} {\frac 2 7a^{\frac {-5} 7}}

=\frac 5 2a^{\frac {-2} {7} + \frac {5} {7}}

=\frac 5 2a^{\frac {3} {7}}

Question 6. \lim_{x \to {\frac {-1} 2}} \frac {8x^3+1} {2x+1}

Solution:

\lim_{x \to {\frac {-1} 2}} \frac {8x^3+1} {2x+1}

=\frac 8 2\lim_{x \to {\frac {-1} 2}} \frac {x^3+{(\frac 1 2)}^3} {x+\frac 1 2}

=4\lim_{x \to {\frac {-1} 2}} \frac {x^3-{(-\frac 1 2)}^3} {x-(-\frac 1 2)}



Applying the formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1} here, n=3   and a=\frac {-1} {2}

=4 \times 3(\frac {-1} 2)^{3-1}

=4 \times 3 \times \frac 1 4

=3

Question 7. \lim_{x \to 27} \frac {({x^{\frac 1 3}+3})({x^{\frac 1 3}-3})} {x-27}

Solution:

\lim_{x \to 27} \frac {({x^{\frac 1 3}+3})({x^{\frac 1 3}-3})} {x-27}

=\lim_{x \to 27} \frac {({x^{\frac 2 3}-9})} {x-27}

=\lim_{x \to 27} \frac {({x^{\frac 2 3}-27^{\frac 2 3}})} {x-27}

Applying the formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}

=\frac 2 3(27)^{\frac 2 3 -1}



=\frac 2 3(27)^{\frac {-1} 3}

=\frac 2 3\times \frac 1 {(27)^{\frac 1 3}}

=\frac 2 3\times \frac 1 3

=\frac 2 9

Question 8. \lim_{x \to 4} \frac {{x^3-64}} {x^2-16}

Solution: 

\lim_{x \to 4} \frac {{x^3-64}} {x^2-16}

=\lim_{x \to 4} \frac {{x^3-4^3}} {x^2-4^2}

Dividing the numerator and denominator with x-4

=\lim_{x \to 4} \frac {\frac {x^3-4^3} {x-4}} {\frac {x^2-4^2} {x-4}}

= \frac {\lim_{x \to 4}{\frac {x^3-4^3} {x-4}}} {\lim_{x \to 4}{\frac {x^2-4^2} {x-4}}}



Applying the formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}here, n=3 in numerator and n=2 in denominator 

= \frac {3(4)^{3-1}} {2(4)^{2-1}}

= \frac {3(4)^{2}} {2(4)}

=6

Question 9. \lim_{x \to 1} \frac {{x^{15}-1}} {x^{10}-1}

Solution

\lim_{x \to 1} \frac {{x^{15}-1}} {x^{10}-1}

Dividing the numerator and denominator with x-1

=\lim_{x \to 1} \frac {\frac {x^{15}-1^{15}} {x-1}} {\frac {x^{10}-1^{10}} {x-1}}

= \frac {\lim_{x \to 1}\frac {x^{15}-1^{15}} {x-1}} {\lim_{x \to 1}\frac {x^{10}-1^{10}} {x-1}}

Applying the formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}here, 



n=15 in numerator and n=10 in denominator

=\frac {15(1)^{15-1}}{10(1)^{10-1}}

=\frac {15} {10}

= \frac 3 2

Question 10. \lim_{x \to {-1}} \frac {{x^{3}+1}} {x+1}

Solution:

\lim_{x \to {-1}} \frac {{x^{3}+1}} {x+1}

=\lim_{x \to {-1}} \frac {{x^{3}-(-1)^3}} {x-(-1)}

Applying the formula \lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}

=3(-1)^{3-1}

=3(-1)^2

=3




My Personal Notes arrow_drop_up
Recommended Articles
Page :