Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.5 | Set 1

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

Summary

Exercise 29.5 Set 1 covers the evaluation of limits of algebraic functions, including polynomials and rational functions, as x approaches infinity or 0, which is crucial for understanding function behavior and calculus, and requires the application of different limit properties and theorems, such as the sum, product, and chain rule, as well as the squeeze theorem, and involves algebraic manipulations, such as factoring and canceling common factors, and the use of L'Hopital's rule for evaluating limits of functions that approach 0/0 or ∞/∞.