Foundations of Limits: Class 11 Mathematics
Welcome to this practice quiz on Limits, a fundamental concept in Calculus for CBSE Class 11. This session focuses on moderate-level problems designed to test your understanding of algebraic manipulations, trigonometric standard limits, and indeterminate forms.
In this quiz, you will encounter:
- Evaluation of limits using factorization and rationalization.
- Application of the standard theorem ##\lim_{x \to a} \frac{x^n – a^n}{x – a} = na^{n-1}##.
- Trigonometric limits involving ##\sin x## and ##\cos x##.
- Exponential and logarithmic limit applications.
Practice these questions to build a strong base for Derivatives and higher calculus topics.
Q1. Evaluate the limit: ##\lim_{x \to 2} \frac{x^2 - 4}{x - 2}##
By factoring the numerator as (x-2)(x+2), the term (x-2) cancels out, leaving x+2. As x approaches 2, the value becomes 2+2=4.
Q2. Find the value of ##\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6}##
Factorize both: ##\frac{(x-3)(x+3)}{(x-3)(x-2)}##. After canceling (x-3), we get ##\frac{x+3}{x-2}##. Substituting x=3 gives 6/1 = 6.
Q3. Evaluate the trigonometric limit: ##\lim_{x \to 0} \frac{\sin 5x}{x}##
Using the standard limit ##\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1##, we multiply and divide by 5 to get ##5 \times \frac{\sin 5x}{5x}##, which results in 5.
Q4. What is the value of ##\lim_{x \to 0} \frac{1 - \cos x}{x^2}##?
Using the identity ##1 - \cos x = 2\sin^2(x/2)##, the limit becomes ##\lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2} = 2 \times (1/2)^2 = 1/2##.
Q5. Evaluate ##\lim_{x \to a} \frac{x^5 - a^5}{x - a}## using standard formulas.
Using the formula ##\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}##, where n=5, the result is ##5a^{5-1} = 5a^4##.
Q6. Find ##\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}## by rationalization.
Multiply numerator and denominator by ##\sqrt{1+x} + 1##. The numerator becomes (1+x)-1 = x. Canceling x gives ##\lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} = 1/2##.
Q7. Evaluate ##\lim_{x \to 0} \frac{\tan 3x}{\sin 2x}##.
Divide numerator and denominator by x: ##\frac{(\tan 3x)/x}{(\sin 2x)/x} = \frac{3 \times 1}{2 \times 1} = 3/2##.
Q8. Find the value of ##\lim_{x \to 2} \frac{x^3 - 8}{x - 2}##.
Factorize ##x^3 - 8## as ##(x-2)(x^2 + 2x + 4)##. Canceling (x-2) leaves ##x^2 + 2x + 4##. Substituting x=2 gives 4+4+4=12.
Q9. If ##\lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 80## and n is a positive integer, find n.
Using ##na^{n-1} = 80##, we have ##n(2)^{n-1} = 80##. For n=5, ##5(2)^4 = 5 \times 16 = 80##.
Q10. Evaluate ##\lim_{x \to 0} \frac{e^{4x} - 1}{x}##.
Using the standard limit ##\lim_{h \to 0} \frac{e^h - 1}{h} = 1##, we multiply and divide by 4 to get ##4 \times \frac{e^{4x} - 1}{4x} = 4##.
Q11. Calculate ##\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}##.
Divide numerator and denominator by (x-1). Apply the formula ##na^{n-1}## to both: ##15(1)^{14} / 10(1)^9 = 15/10 = 1.5##.
Q12. Find ##\lim_{x \to 0} \frac{\sin^2 3x}{x^2}##.
The expression can be written as ##(\frac{\sin 3x}{x})^2 = (3 \times \frac{\sin 3x}{3x})^2 = 3^2 = 9##.
Q13. Evaluate ##\lim_{x \to 0} \frac{\log(1 + 2x)}{x}##.
Using the standard limit ##\lim_{x \to 0} \frac{\log(1+x)}{x} = 1##, we adjust for 2x to get ##2 \times \frac{\log(1+2x)}{2x} = 2 \times 1 = 2##.
Q14. What is ##\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}##?
Using ##1 - \cos 4x = 2\sin^2 2x##, the limit is ##\lim_{x \to 0} \frac{2\sin^2 2x}{x^2} = 2 \times 2^2 = 8##.
Q15. Evaluate ##\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}##.
Rationalize by multiplying with ##\sqrt{1+x} + \sqrt{1-x}##. Numerator becomes (1+x)-(1-x)=2x. Canceling x leaves ##2/(\sqrt{1+x} + \sqrt{1-x})##. As x approaches 0, it becomes 2/2=1.
Q16. Find ##\lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3}##.
Factorize the numerator: ##(x-3)(x+1)##. Canceling (x-3) leaves x+1. Substituting x=3 gives 3+1=4.
Q17. Evaluate ##\lim_{x \to \pi} \frac{\sin x}{\pi - x}##.
Let ##y = \pi - x##. As ##x \to \pi, y \to 0##. Also ##\sin x = \sin(\pi - y) = \sin y##. The limit becomes ##\lim_{y \to 0} \frac{\sin y}{y} = 1##.
Q18. Evaluate ##\lim_{x \to 0} \frac{\sin 2x + 3x}{2x + \sin 3x}##.
Divide every term by x: ##\frac{(\sin 2x)/x + 3}{2 + (\sin 3x)/x} = \frac{2 + 3}{2 + 3} = 5/5 = 1##.
Q19. Find ##\lim_{x \to 0} \frac{1 - \cos 6x}{1 - \cos 4x}##.
Using the formula ##\lim_{x \to 0} \frac{1 - \cos ax}{x^2} = a^2/2##, the limit is ##(6^2/2) / (4^2/2) = 36/16 = 9/4##.
Q20. Calculate ##\lim_{x \to 1} \frac{x^4 - 1}{x^3 - 1}##.
Divide numerator and denominator by (x-1). Apply ##na^{n-1}##: Numerator limit is 4, denominator limit is 3. Result is 4/3.
Q21. Evaluate ##\lim_{x \to 0} \frac{\tan 2x - x}{3x - \sin x}##.
Divide by x: ##\frac{(\tan 2x)/x - 1}{3 - (\sin x)/x} = \frac{2 - 1}{3 - 1} = 1/2##.
Q22. Find ##\lim_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a}##.
Using ##x^{1/2} - a^{1/2}## formula, the result is ##(1/2)a^{(1/2)-1} = (1/2)a^{-1/2} = 1/(2\sqrt{a})##.
Q23. Evaluate ##\lim_{x \to 0} \frac{7^x - 1}{x}##.
Using the standard exponential limit formula ##\lim_{x \to 0} \frac{a^x - 1}{x} = \log a##, the value is ##\log 7##.
Q24. Find ##\lim_{x \to 0} \frac{\sin ax}{\tan bx}##.
Divide numerator and denominator by x: ##\frac{(\sin ax)/x}{(\tan bx)/x} = a/b##.
Q25. Evaluate ##\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}##.
Using the power formula with n=1/2 and a=1: ##(1/2)(1)^{(1/2)-1} = 1/2##.
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