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Class 11 RD Sharma Solutions- Chapter 29 Limits – Exercise 29.2

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  • Last Updated : 02 Feb, 2021
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Question 1. Evaluate  \lim_{x \to 1} \frac {x^2+1} {x+1}

Solution:

\lim_{x \to 1} \frac {x^2+1} {x+1}=\frac {1^2+1} {1+1}=\frac 2 2 = 1

Question 2. Evaluate \lim_{x \to 0} \frac {2x^2+3x+4} {x^2+3x+2}

Solution:

\lim_{x \to 0} \frac {2x^2+3x+4} {x^2+3x+2}= \frac {2(0)^2+3(0)+4} {(0)^2+3(0)+2}= \frac 4 2=2

Question 3. Evaluate \lim_{x \to 3} \frac {\sqrt{2x+3}} {x+3}

Solution:

\lim_{x \to 3} \frac {\sqrt{2x+3}} {x+3}=\frac {\sqrt{2(3)+3}} {3+3}=\frac {\sqrt{9}} {6}=\frac 3 6=\frac 1 2

Question 4. Evaluate \lim_{x \to 1} \frac {\sqrt{x+8}} {\sqrt{x}}

Solution: 

\lim_{x \to 1} \frac {\sqrt{x+8}} {\sqrt{x}}=\frac {\sqrt{1+8}} {\sqrt{1}}=\frac {3} {1}=3

Question 5. Evaluate \lim_{x \to a} \frac {\sqrt{x} + \sqrt{a}} {x+a}

Solution:

\lim_{x \to a} \frac {\sqrt{x} + \sqrt{a}} {x+a}=\frac {\sqrt{a} + \sqrt{a}} {a+a}=\frac {2\sqrt{a}} {2a}=\frac 1 {\sqrt{a}}

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

Solution:

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

Question 7. Evaluate \lim_{x \to 0} \frac {x^{\frac 2 3}-9} {x-27}

Solution:

\lim_{x \to 0} \frac {x^{\frac 2 3}-9} {x-27}=\frac {(0)^{\frac 2 3}-9} {0-27}=\frac {-9}{-27}=\frac 1 3

Question 8. Evaluate \lim_{x \to 0} 9

Solution:

\lim_{x \to 0} 9=9

Question 9. Evaluate \lim_{x \to 2} 3-x

Solution:

\lim_{x \to 2} 3-x=3-2=1

Question 10. Evaluate \lim_{x \to -1} 4x^2+2

Solution:

\lim_{x \to -1} 4x^2+2=4(-1)^2+2=4+2=6

Question 11.  Evaluate  \lim_{x \to -1} \frac {x^3-3x+1} {x-1}

Solution:

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

Question 12. Evaluate \lim_{x \to 0} \frac {3x+1} {x+3}

Solution:

\lim_{x \to 0} \frac {3x+1} {x+3}=\frac {3(0)+1} {0+3}=\frac 1 3

Question 13. Evaluate \lim_{x \to 3} \frac {x^2-9} {x+2}

Solution:

\lim_{x \to 3} \frac {x^2-9} {x+2}=\frac {3^2-9} {3+2}=\frac 0 5=0

Question 14. Evaluate \lim_{x \to 0} \frac {ax+b} {x+d}

Solution:

\lim_{x \to 0} \frac {ax+b} {x+d}=\frac {a(0)+b} {0+d}=\frac b d


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