(D) For $x \to 0^+$,$[x] = 0$ and $\{x\} = x$. Thus,$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\tan^2 x}{x^2 - 0^2} = \lim_{x \to 0^+} \left( \fr

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(D) For $x \to 0^+$,$[x] = 0$ and $\{x\} = x$. Thus,$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\tan^2 x}{x^2 - 0^2} = \lim_{x \to 0^+} \left( \frac{\tan x}{x} \right)^2 = 1^2 = 1$.For $x \to 0^-$,let $x = -h$ where $h > 0$. Then $[x] = -1$ and $\{x\} = x - [x] = -h - (-1) = 1 - h$.$\lim_{x \to 0^-} f(x) = \lim_{h \to 0^+} \sqrt{(1-h) \cot(1-h)} = \sqrt{1 \cdot \cot(1)} = \sqrt{\cot 1}$.Now,$\left( \lim_{x \to 0^-} f(x) \right)^2 = (\sqrt{\cot 1})^2 = \cot 1$.Therefore,$\cot^{-1} \left( \lim_{x \to 0^-} f(x) \right)^2 = \cot^{-1}(\cot 1) = 1$.Thus,both $(A)$ and $(C)$ are correct.

Class 12 Mathematics Continuity and Differentiation Continuity

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 greatest integer function and $\{x\}$ is the fractional part function of $x$,then:

A
$\lim_{x \to 0^+} f(x) = 1$
B
$\lim_{x \to 0^-} f(x) = 1$
C
$\cot^{-1} \left( \lim_{x \to 0^-} f(x) \right)^2 = 1$
D
Both $(A)$ and $(C)$

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If the function defined by $f(x) = \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x}, x \neq 0$ is continuous at $x=0$,then $6 f(0)$ is equal to

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Let $f$ and $g$ be real-valued functions. If $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{[f(x)+7]^{2 / 3}}=\frac{7}{4}$, $\lim _{x \rightarrow 0} f(x)=1$ and $\lim _{x \rightarrow 0} g(x)=\alpha$, then $h(x)= \begin{cases} \sin (\alpha x), & 0 \leq x \leq \frac{\pi}{10} \\ \cos (2 \alpha x), & \frac{\pi}{10} < x \leq \frac{\pi}{5} \end{cases}$ is:

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If $f(x) = \begin{cases} \frac{a + 3\cos x}{x^2}, & x < 0 \\ b\tan \left( \frac{\pi}{[x + 3]} \right), & x \geqslant 0 \end{cases}$ is continuous at $x = 0$,then:

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If the function $f(x)=\begin{cases} \frac{1-\cos x}{x^{2}}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then the value of $k$ is

EasyKCET 2007

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Let a function $f: R \rightarrow R$ be defined as
$f(x) = \begin{cases} \sin x - e^x & \text{if } x \leq 0 \\ a + [-x] & \text{if } 0 < x < 1 \\ 2x - b & \text{if } x \geq 1 \end{cases}$
where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous on $R$,then $(a + b)$ is equal to:

DifficultJEE MAIN 2021

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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 greatest integer function and $\{x\}$ is the fractional part function of $x$,then:

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If the function defined by $f(x) = \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x}, x \neq 0$ is continuous at $x=0$,then $6 f(0)$ is equal to

If $f(x) = \begin{cases} \frac{a + 3\cos x}{x^2}, & x < 0 \\ b\tan \left( \frac{\pi}{[x + 3]} \right), & x \geqslant 0 \end{cases}$ is continuous at $x = 0$,then:

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