Mastering Limits: Class 11 Mathematics Quiz
This quiz is designed for students beginning their journey into Calculus. It focuses on the fundamental concept of Limits as per the CBSE Class 11 syllabus. Evaluating limits is a crucial skill for understanding continuity and derivatives later in the course.
In this practice set, you will be tested on:
- The algebra of limits and direct substitution.
- Methods of factorization and rationalization to resolve indeterminate forms.
- Application of standard trigonometric limits like
##\lim_{x \to 0} \frac{\sin x}{x} = 1##. - Understanding left-hand limits (LHL) and right-hand limits (RHL).
- Exponential and logarithmic limit formulas.
Difficulty Level: Moderate
Q1. What is the value of ##\lim_{x \to 3} \frac{x^2 - 9}{x - 3}##?
Factor the numerator as ##(x-3)(x+3)##. The expression becomes ##\frac{(x-3)(x+3)}{x-3} = x+3##. Substituting ##x=3## gives ##3+3=6##.
Q2. Evaluate the limit: ##\lim_{x \to 0} \frac{\sin 3x}{x}##.
Multiply and divide by 3 to get ##3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x}##. Since ##\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1##, the result is ##3 \cdot 1 = 3##.
Q3. Find the value of ##\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}##.
Using the formula ##\lim_{x \to a} \frac{x^n - a^n}{x^m - a^m} = \frac{n}{m} a^{n-m}##, we get ##\frac{15}{10} \cdot 1^{15-10} = 1.5##, which simplifies to ##3/2##.
Q4. Evaluate ##\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}##.
Rationalize the numerator by multiplying by ##\sqrt{1+x} + 1##. The expression becomes ##\frac{(1+x)-1}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}##. Substituting ##x=0## gives ##1/2##.
Q5. For a function ##f(x)##, if the left-hand limit is 3 and the right-hand limit is 3 at ##x = a##, then ##\lim_{x \to a} f(x)## is:
A limit exists at a point if and only if the left-hand limit and the right-hand limit are equal. Since both are 3, the limit is 3.
Q6. What is the value of ##\lim_{x \to 0} \frac{\tan x}{x}##?
Using the identity ##\tan x = \sin x / \cos x##, the limit is ##\lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot 1 = 1##.
Q7. Evaluate ##\lim_{x \to 1} \frac{x^3 - 1}{x - 1}##.
Factor the numerator as ##(x-1)(x^2 + x + 1)##. The expression simplifies to ##x^2 + x + 1##. Substituting ##x=1## gives ##1^2 + 1 + 1 = 3##.
Q8. Find the value of ##\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}##.
Using the identity ##1 - \cos 2x = 2\sin^2 x##, the limit becomes ##\lim_{x \to 0} \frac{2\sin^2 x}{x^2} = 2 (\lim_{x \to 0} \frac{\sin x}{x})^2 = 2(1)^2 = 2##.
Q9. According to the standard formula, ##\lim_{x \to a} \frac{x^n - a^n}{x - a}## is equal to:
This is a standard theorem in limits where the derivative of ##x^n## at ##x=a## is derived using the limit definition.
Q10. Evaluate ##\lim_{x \to 2} \frac{x^3 - 8}{x - 2}##.
Factor the numerator as ##(x-2)(x^2 + 2x + 4)##. The expression simplifies to ##x^2 + 2x + 4##. Substituting ##x=2## gives ##4 + 4 + 4 = 12##.
Q11. What is the limit of a constant function ##f(x) = k## as ##x## approaches any value ##a##?
The value of a constant function does not change regardless of what ##x## approaches, so the limit is always the constant itself.
Q12. Evaluate ##\lim_{x \to 0} \frac{e^{5x} - 1}{x}##.
Using the standard limit ##\lim_{x \to 0} \frac{e^x - 1}{x} = 1##, we multiply and divide by 5 to get ##5 \cdot \lim_{x \to 0} \frac{e^{5x} - 1}{5x} = 5 \cdot 1 = 5##.
Q13. Find ##\lim_{x \to \infty} \frac{2x + 3}{5x - 1}##.
Divide the numerator and denominator by ##x## to get ##\frac{2 + 3/x}{5 - 1/x}##. As ##x \to \infty##, ##3/x## and ##1/x## approach 0, leaving ##2/5##.
Q14. Evaluate ##\lim_{x \to 0} \frac{\sin 5x}{\sin 2x}##.
Multiply and divide the numerator by ##5x## and the denominator by ##2x##. The limit becomes ##\frac{(\sin 5x / 5x) \cdot 5x}{(\sin 2x / 2x) \cdot 2x} = 5/2##.
Q15. What is the value of the left-hand limit ##\lim_{x \to 0^-} \frac{|x|}{x}##?
For ##x < 0##, ##|x| = -x##. Therefore, the expression is ##-x/x = -1##. The limit as ##x## approaches 0 from the left is -1.
Q16. Find ##\lim_{x \to 0} \frac{\log(1+2x)}{x}##.
Using the standard limit ##\lim_{x \to 0} \frac{\log(1+x)}{x} = 1##, we multiply and divide by 2 to get ##2 \cdot \lim_{x \to 0} \frac{\log(1+2x)}{2x} = 2 \cdot 1 = 2##.
Q17. Evaluate ##\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}##.
This can be written as ##\lim_{x \to 1} \frac{x^{1/2} - 1^{1/2}}{x - 1}##. Applying the formula ##na^{n-1}##, we get ##(1/2)(1)^{-1/2} = 1/2##.
Q18. Find the value of ##\lim_{x \to 0} \frac{1 - \cos x}{x^2}##.
Using ##1 - \cos x = 2\sin^2(x/2)##, the limit is ##\lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2} = 2 \cdot \frac{1}{4} \cdot \lim_{x \to 0} (\frac{\sin(x/2)}{x/2})^2 = 1/2##.
Q19. Evaluate ##\lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2}##.
Factor the numerator as ##(x+2)(x+3)##. The expression simplifies to ##x+3##. Substituting ##x=-2## gives ##-2+3=1##.
Q20. Evaluate ##\lim_{x \to 0} \frac{\sin ax}{bx}## where ##a, b \neq 0##.
Rewrite the expression as ##\frac{a}{b} \cdot \frac{\sin ax}{ax}##. The limit as ##x \to 0## is ##a/b \cdot 1 = a/b##.
Q21. Evaluate the limit ##\lim_{x \to 2^+} [x]##, where ##[x]## denotes the greatest integer function.
As ##x## approaches 2 from the right (##x > 2##), the greatest integer less than or equal to ##x## is 2. Thus the RHL is 2.
Q22. Evaluate ##\lim_{x \to \pi/2} \frac{\cos x}{\pi/2 - x}##.
Let ##y = \pi/2 - x##. As ##x \to \pi/2##, ##y \to 0##. Since ##\cos x = \sin(\pi/2 - x) = \sin y##, the limit becomes ##\lim_{y \to 0} \frac{\sin y}{y} = 1##.
Q23. What is the value of ##\lim_{x \to 0} \frac{a^x - 1}{x}## for ##a > 0##?
This is a standard exponential limit result derived from the definition of the derivative of ##a^x## at ##x=0##.
Q24. Evaluate ##\lim_{x \to 1} \frac{x^7 - 1}{x - 1}##.
Applying the formula ##\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}## with ##n=7## and ##a=1##, we get ##7(1)^6 = 7##.
Q25. Find ##\lim_{x \to 0} \frac{\tan 2x}{\sin 3x}##.
Multiply and divide by ##2x## and ##3x## respectively: ##\frac{(\tan 2x / 2x) \cdot 2x}{(\sin 3x / 3x) \cdot 3x} = \frac{1 \cdot 2x}{1 \cdot 3x} = 2/3##.
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