Let $\mathrm{f}, \mathrm{g}$ and h be the real valued functions defined on $\mathbb{R}$ as $f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$, $\mathrm{g}(\mathrm{x})=\left\{\begin{array}{cl}\frac{\sin (\mathrm{x}+1)}{(\mathrm{x}+1)}, & \mathrm{x} \neq-1 \\ 1, & \mathrm{x}=-1\end{array}\right.$ and $\mathrm{h}(\mathrm{x})=2[\mathrm{x}]-\mathrm{f}(\mathrm{x})$, where $[x]$ is the greatest integer $\leq x$. Then the value of $\lim _{\mathrm{x} \rightarrow 1} \mathrm{~g}(\mathrm{~h}(\mathrm{x}-1))$ is

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Limits

Let

f, g

and h be the real valued functions defined on

R

f (x) = {\frac{x}{∣ x ∣}, 1, x \neq = 0 x = 0

g (x) = {\frac{s i n ( x + 1 )}{( x + 1 )}, 1, x \neq = - 1 x = - 1

and

h (x) = 2 [x] - f (x)

,
where

[x]

is the greatest integer

\leq x

. Then the value of

lim_{x \to 1} g (h (x - 1))

Answer: (A)

Put

u = x - 1

, so

u \to 0

x \to 1

. For

x

just below

1

u \to 0^{-}

gives

[u] = - 1

and

f (u) = \frac{u}{∣ u ∣} = - 1

, hence

h (u) = 2 (- 1) - (- 1) = - 1

. For

x

just above

1

u \to 0^{+}

gives

[u] = 0

and

f (u) = 1

, hence

h (u) = 2 (0) - 1 = - 1

. On both sides

h (x - 1) = - 1

, and

g

- 1

takes its defined value

g (- 1) = 1

. The limit equals

1

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option