Continuity
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision.
Continuity — Key Facts for CUET • Essential Formula: $\lim_{x \to a} f(x) = f(a)$ (function must equal its limit at the point) • Most Tested Concept: Checking continuity at a point using the three-condition test (left limit = right limit = function value) • Common Mistake: Students forget that continuity requires the function to be defined at the point, not just have matching limits • Key Technique: For piecewise functions, always check continuity at the boundary point by evaluating both pieces • Important Exception: Rational functions are continuous everywhere except where denominator = 0 (very frequent CUET trap!) • Frequent Question Type: Find ‘k’ so that function is continuous at x = a (parameter finding problems) ⚡ Exam Tip: When asked to find continuity of composite functions at corners, ALWAYS check the inner function’s behavior first, then apply the definition to the outer function.
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Continuity — CUET Study Guide
A function f(x) is continuous at x = a if and only if three conditions hold simultaneously: $\lim_{x \to a} f(x)$ exists, $f(a)$ exists, and $\lim_{x \to a} f(x) = f(a)$. This three-part test is the foundation for all continuity problems in CUET. Understanding this definition helps avoid the most common error—checking only limits without verifying the function’s value at the point.
Standard Forms: Polynomials are continuous everywhere. Rational functions $\frac{p(x)}{q(x)}$ are continuous where $q(x) \neq 0$. Trigonometric functions like sin(x) and cos(x) are continuous on their domains, while tan(x) and sec(x) have discontinuities where cos(x) = 0.
Typical CUET Patterns: (1) Finding the value of a constant to make a piecewise function continuous, (2) Determining points of discontinuity for rational/trigonometric functions, (3) Using continuity to evaluate limits by direct substitution.
Common Traps: Students often forget to check if the function is defined at the point. Also, when dealing with modulus functions $|f(x)|$, remember that continuity depends on $f(x)$ being continuous.
Example 1: Find k if $f(x) = \begin{cases} kx + 3 & x < 2 \ x^2 - 1 & x \geq 2 \end{cases}$ is continuous at x = 2. Solution: For continuity at x = 2: $\lim_{x \to 2^-} (kx + 3) = 2k + 3$ must equal $f(2) = 4 - 1 = 3$. Thus $2k + 3 = 3$, giving $k = 0$.
Example 2: Discuss continuity of $f(x) = \frac{x^2 - 4}{x - 2}$ at x = 2. Solution: Simplify: $\frac{(x-2)(x+2)}{x-2} = x + 2 Content adapted based on your selected roadmap duration. Switch tiers using the pill selector above.