Limits, Continuity & Differentiability
Continuity And Differentiability
nta_abhyas_2025
Grade 12
Question:
If $f(x) = \begin{cases} \frac{\sqrt{x+3}-2}{x-1} & 0 \leq x < 4 \\ b & x \geq 4 \end{cases}$ is continuous at $x = 4$, then value of $\frac{a}{b}$ is equal to
Step-by-Step Solution
Key Concept: Rationalization of square root expressions using conjugate multiplication is essential for resolving indeterminate forms.
For $f(x)$ to be continuous at $x = 4$, we need $\lim_{x \to 4} f(x) = f(4) = b$. The numerator requires rationalization: for the expression $\sqrt{1+x+x^2}$ near $x = 4$, multiply by the conjugate. After simplification using $a = 2$ and rationalizing: $\lim_{x \to 4} \frac{(1+x+x^2)-1}{(x-4)(\sqrt{1+x+x^2}+1)} = \lim_{x \to 4} \frac{x(1+x)}{(x-4)(\text{conjugate})}$. Through careful algebraic manipulation and using $a = 2$, we find $b = 12$.
Correct Answer: 12