Let f , g and h be real-valued functions defined on R f ( x ) = { x | x | , x 0 1 , x = 0 , g ( x ) = { sin ( x + 1 ) ( x + 1 ) , x- 1 1 , x = - 1 and h ( x ) = 2 [ x ] - f ( x ) , where [ x ] is the greatest integer x . Then the value of

Question

$Let f, g and h be real-valued functions defined on R$ $f (x) =$ ${\begin{matrix} \frac{x}{| x |}, & x \neq 0 \\ 1, & x = 0 \end{matrix},$ $g (x) =$ ${\begin{matrix} \frac{\sin (x + 1)}{(x + 1)}, & x \neq - 1 \\ 1, & x = - 1 \end{matrix}$ and $h (x) = 2 [x] - f (x),$ where $[x]$ is the greatest integer $\leq x$ . Then the value of $\lim_{x \to 1} g (h (x - 1))$ is [2023]

Smart Achievers · Accepted Answer

(1)

$f (x) =$ ${\begin{matrix} \frac{x}{| x |}, & x \neq 0 \\ 1, & x = 0 \end{matrix}$

and $g (x) =$ ${\begin{matrix} \frac{\sin (x + 1)}{x + 1}, & x \neq - 1 \\ 1, & x = - 1 \end{matrix}$

$h (x) = 2 [x] - f (x), [x]$ is the greatest integer $\leq x$

$LHL$ $= \lim_{x \to 1^{-}} g (h (x - 1)) = \lim_{k \to 0} g (h (1 - k - 1))$

$= \lim_{k \to 0} g [(2 [- k] - \frac{- k}{| - k |})] = g [2 (- 1) + 1] = \lim_{x \to 1^{-}} g (- 1) = 1$

$RHL$ $= \lim_{x \to 1^{+}} g (h (x - 1)) = \lim_{h \to 0} g (h (1 + k - 1))$

$= \lim_{k \to 0} g (2 [k] - f (k)) = \lim_{h \to 0} g (0 - 1) = \lim_{x \to 1^{+}} g (- 1) = 1$

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

Solution

1 1

2 0

3 sin (1)

4 - 1

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Solution

Q. Let f,g and h be real-valued functions defined on R f(x)={x|x|,x≠01,x=0, g(x)={sin(x+1)(x+1),x≠-11,x=-1 and h(x)=2[x]-f(x), where [x] is the greatest integer ≤x. Then the value of limx→1g(h(x-1)) is [2023]

1 1

2 0

3 sin (1)

4 - 1

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