Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(x) = \begin{cases} \frac{\tan^2 [x]}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{[x]} \cot [x] & \text{for } x < 0 \end{cases}$ where $[x]$ is the step up function and $\{x\}$ is the fractional part function of $x$, then :
$\lim_{x \to 0^+} f(x) = 1$
$\lim_{x \to 0^-} f(x) = 1$
$\cot^{-1} \left[ \lim_{x \to 0^-} f(x) \right]^2 = 1$
$f$ is continuous at $x = 1$
Step-by-Step Solution
Key Concept: Strategic substitution and standard limit formulas like $\lim_{u \to 0} \frac{\tan(u)}{u} = 1$ simplify complex limit expressions.
For the first limit, substitute $h = \tan(x)$ so as $x \to 0$, $h \to 0$, giving $\lim_{x \to 0} \frac{\tan^2(x)}{x^2 - [x]^2} = \lim_{h \to 0} \frac{\tan^2(h)}{h^2} = 1$ where $x > 0$. For the second limit, compute $\lim_{x \to 0} \sqrt{x} \cot(x) = \lim_{h \to 0} \sqrt{\frac{1-h}{\tan(1-h)}} = \sqrt{\cot(1)}$ by substitution.
Correct Answer: 1,3