JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Limit of a Composite Piecewise Function

Q: Why do we evaluate g(-1) directly rather than taking a limit of sin(x+1)/(x+1) as x → -1?

Because for all sufficiently small non-zero δ, h(±δ) evaluates to the exact constant value -1. It does not merely approach -1, so g(h(u)) is identically g(-1) = 1 in a deleted neighborhood.

Let f, g and h be the real valued functions defined on ℝ as 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 lim as x → 1 of g(h(x - 1)) is

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What feels right?

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A-1B0Csin(1)D1

Step-by-step solutionView

Correct answer

As $x \to 1$ from both sides, the inner expression $h (x - 1)$ evaluates to $- 1$ , so $g (h (x - 1))$ approaches $g (- 1) = 1$ .

Option analysis

Why each option works or fails

A · -1

Believing that the question asks for the limit of the inner argument $h (x - 1)$ rather than the composition $g (h (x - 1))$ . Evaluate the outer function $g$ at the value obtained from the inner function, noticing $g (- 1) = 1$ .

B · 0

Incorrectly evaluating $\sin (x + 1)$ or assuming the limit of $g (t)$ as $t \to 0$ applies instead of $t = - 1$ . Check the value of $h (x - 1)$ ; since $h (x - 1) = - 1$ , substitute $- 1$ into $g (x)$ which yields the defined value $1$ , not $0$ .

C · sin(1)

Assuming the argument inside $g$ approaches $0$ and using the non-branch formula $\frac{\sin (0 + 1)}{0 + 1} = \sin (1)$ . Compute the inner function carefully: $h (x - 1)$ equals $- 1$ in a neighborhood of $x = 1$ (for $x \neq 1$ ), and $g (- 1) = 1$ by definition.

D · 1

This correctly evaluates the left- and right-hand limits of $h (x - 1)$ to find that $h (x - 1) = - 1$ for all $x$ in a deleted neighborhood of $1$ , giving $g (- 1) = 1$ . Correctly observed that for both $x \in (0, 1)$ and $x \in (1, 2)$ , $h (x - 1) = - 1$ , hence $g (h (x - 1)) = g (- 1) = 1$ .

Reviewed route

Solution

StepWorking

01Given

$f (x) = {\begin{cases} \frac{x}{∣ x ∣}, & x \neq 0 \\ 1, & x = 0 \end{cases}$ , $g (x) = {\begin{cases} \frac{\sin (x + 1)}{x + 1}, & x \neq - 1 \\ 1, & x = - 1 \end{cases}$ , and $h (x) = 2 [x] - f (x)$ . We seek $\lim_{x \to 1} g (h (x - 1))$ .

02Approach

Substitute $u = x - 1$ , so as $x \to 1$ , $u \to 0$ . We evaluate the left-hand limit ( $u \to 0^{-}$ ) and right-hand limit ( $u \to 0^{+}$ ) of $g (h (u))$ .

03Execute

For $u \to 0^{-}$ (let $u = - δ$ , $δ > 0$ small): $[- δ] = - 1$ and $f (- δ) = \frac{- δ}{∣ - δ ∣} = - 1$ . Thus $h (- δ) = 2 (- 1) - (- 1) = - 2 + 1 = - 1$ . Then $g (h (- δ)) = g (- 1) = 1$ .

04Execute

For $u \to 0^{+}$ (let $u = δ$ , $δ > 0$ small): $[δ] = 0$ and $f (δ) = \frac{δ}{∣ δ ∣} = 1$ . Thus $h (δ) = 2 (0) - 1 = - 1$ . Then $g (h (δ)) = g (- 1) = 1$ .

✓Verify

Both $LHL = 1$ and $RHL = 1$ . Since both one-sided limits are equal, $\lim_{x \to 1} g (h (x - 1)) = 1$ .

Hints that build this answer step by step

To find $\lim_{x \to 1} g (h (x - 1))$ , what is the most reliable strategy?
Evaluate the one-sided limits of $h (x - 1)$ as $x \to 1^{+}$ and $x \to 1^{-}$ separately.
For $x \in (1, 2)$ , let $t = x - 1 \in (0, 1)$ . What is the exact value of $h (t) = 2 [t] - f (t)$ ?
$- 1$
For $x \in (0, 1)$ , let $t = x - 1 \in (- 1, 0)$ . What is the exact value of $h (t) = 2 [t] - f (t)$ ?
$- 1$
Since $h (x - 1) = - 1$ identically for all $x \neq 1$ in $(0, 2)$ , what is $\lim_{x \to 1} g (h (x - 1))$ ?
$1$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we evaluate

g (- 1)

directly rather than taking a limit of

\sin (x + 1) / (x + 1)

x \to - 1

Because for all sufficiently small non-zero $δ$ , $h (\pm δ)$ evaluates to the exact constant value $- 1$ . It does not merely approach $- 1$ , so $g (h (u))$ is identically $g (- 1) = 1$ in a deleted neighborhood.

Answer

As $x \to 1$ from both sides, the inner expression $h (x - 1)$ evaluates to $- 1$ , so $g (h (x - 1))$ approaches $g (- 1) = 1$ .

Why each option works or fails

A: -1 - Believing that the question asks for the limit of the inner argument $h (x - 1)$ rather than the composition $g (h (x - 1))$ . Evaluate the outer function $g$ at the value obtained from the inner function, noticing $g (- 1) = 1$ .
B: 0 - Incorrectly evaluating $\sin (x + 1)$ or assuming the limit of $g (t)$ as $t \to 0$ applies instead of $t = - 1$ . Check the value of $h (x - 1)$ ; since $h (x - 1) = - 1$ , substitute $- 1$ into $g (x)$ which yields the defined value $x \to 1$ , not $0$ .
C: sin(1) - Assuming the argument inside $g$ approaches $0$ and using the non-branch formula $\frac{\sin (0 + 1)}{0 + 1} = \sin (1)$ . Compute the inner function carefully: $h (x - 1)$ equals $- 1$ in a neighborhood of $x = 1$ (for $x \neq 1$ ), and $g (- 1) = 1$ by definition.
D · correct: 1 - This correctly evaluates the left- and right-hand limits of $h (x - 1)$ to find that $h (x - 1) = - 1$ for all $x$ in a deleted neighborhood of $1$ , giving $g (- 1) = 1$ . Correctly observed that for both $x \in (0, 1)$ and $x \in (1, 2)$ , $h (x - 1) = - 1$ , hence $g (h (x - 1)) = g (- 1) = 1$ .

Step-by-step solution

Given: $f (x) = {\begin{cases} \frac{x}{∣ x ∣}, & x \neq 0 \\ 1, & x = 0 \end{cases}$ , $g (x) = {\begin{cases} \frac{\sin (x + 1)}{x + 1}, & x \neq - 1 \\ 1, & x = - 1 \end{cases}$ , and $h (x) = 2 [x] - f (x)$ . We seek $\lim_{x \to 1} g (h (x - 1))$ .
Approach: Substitute $u = x - 1$ , so as $x \to 1$ , $u \to 0$ . We evaluate the left-hand limit ( $u \to 0^{-}$ ) and right-hand limit ( $u \to 0^{+}$ ) of $g (h (u))$ .
Execute: For $u \to 0^{-}$ (let $u = - δ$ , $δ > 0$ small): $[- δ] = - 1$ and $f (- δ) = \frac{- δ}{∣ - δ ∣} = - 1$ . Thus $h (- δ) = 2 (- 1) - (- 1) = - 2 + 1 = - 1$ . Then $g (h (- δ)) = g (- 1) = 1$ .
Execute: For $u \to 0^{+}$ (let $u = δ$ , $δ > 0$ small): $[δ] = 0$ and $f (δ) = \frac{δ}{∣ δ ∣} = 1$ . Thus $h (δ) = 2 (0) - 1 = - 1$ . Then $g (h (δ)) = g (- 1) = 1$ .
Verify: Both $LHL = 1$ and $RHL = 1$ . Since both one-sided limits are equal, $\lim_{x \to 1} g (h (x - 1)) = 1$ .

Limit of a Composite Piecewise Function

What feels right?

As $x \to 1$ from both sides, the inner expression $h (x - 1)$ evaluates to $- 1$ , so $g (h (x - 1))$ approaches $g (- 1) = 1$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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