Why do we evaluate directly rather than taking a limit of as ?

Because for all sufficiently small non-zero , evaluates to the exact constant value . It does not merely approach , so is identically in a deleted neighborhood.

JEE MainMathematics

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Limits, Continuity and Differentiability: Mathematics | JEE Main

Let

\text{f}, \text{g}

and

\text{h}

be the real valued functions defined on

\mathbb{R}

\text{f}(\text{x}) = \begin{cases} \frac{\text{x}}{|\text{x}|}, & \text{x} \ne 0 \\ 1, & \text{x} = 0 \end{cases}, \text{g}(\text{x}) = \begin{cases} \frac{\sin(\text{x} + 1)}{(\text{x} + 1)}, & \text{x} \ne -1 \\ 1, & \text{x} = -1 \end{cases}

and

\text{h}(\text{x}) = 2[\text{x}] - \text{f}(\text{x})

, where

[\text{x}]

is the greatest integer

\le \text{x}

. Then the value of

\lim_{\text{x} \rightarrow 1}\text{g}(\text{h}(\text{x} - 1))

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

-1

0

\sin(1)

1

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 $x \ne 1$ ), and $g(-1) = 1$ by definition.

D ·

1

None. 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 $u = -\delta$ , $\delta > 0$ small): $[- \delta] = -1$ and $f(-\delta) = \frac{-\delta}{|-\delta|} = -1$ . Thus $h(-\delta) = 2(-1) - (-1) = -2 + 1 = -1$ . Then $g(h(-\delta)) = g(-1) = 1$ .

04execute

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

✓verify

Both $\text{LHL} = 1$ and $\text{RHL} = 1$ . 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 $x \ne 1$ 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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\delta$ , $h(\pm\delta)$ 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 $1$ , not $0$ .
C: $\sin(1)$ - Assuming the argument inside $[\text{x}]$ 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 $x \ne 1$ ), and $g(-1) = 1$ by definition.
D · correct: $1$ - None. 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 $u = -\delta$ , $\delta > 0$ small): $[- \delta] = -1$ and $f(-\delta) = \frac{-\delta}{|-\delta|} = -1$ . Thus $h(-\delta) = 2(-1) - (-1) = -2 + 1 = -1$ . Then $g(h(-\delta)) = g(-1) = 1$ .
execute: For $u \to 0^+$ (let $u = \delta$ , $\delta > 0$ small): $[\delta] = 0$ and $f(\delta) = \frac{\delta}{|\delta|} = 1$ . Thus $h(\delta) = 2(0) - 1 = -1$ . Then $g(h(\delta)) = g(-1) = 1$ .
verify: Both $\text{LHL} = 1$ and $\text{RHL} = 1$ . Since both one-sided limits are equal, $\lim_{x \to 1} g(h(x-1)) = 1$ .

Limits, Continuity and Differentiability: Mathematics | JEE Main

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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