Let f, g and h be the real valued functions defined on R as f(x) = { x/|x| for x ≠ 0, 1 for x = 0 },

Q: Wait, why can't I just plug $x=1$ into the function immediately? Isn't that the same as taking the limit?

Great question! In JEE Advanced, we must distinguish between the value of a function at a point and the limit as we approach that point. While sometimes they are the same (for continuous functions), composite functions can be tricky. If the inner function is discontinuous, direct substitution can lead to catastrophic errors. Always evaluate the limit of the inner function first!

Q: Why does the problem define $f(0) = 1$? I thought limits don't care about the value at the point.

You are absolutely right that limits don't care about the value at the point! However, in this specific problem, we are evaluating $g(h(0))$. Since $h(0)$ involves $f(0)$, we need that definition to evaluate the function at the specific point $t=0$. It's a test of whether you can handle piecewise definitions rigorously.

Q: How can $h(t)$ be constant? It has $[t]$ in it, which is a step function!

That is the beauty of this problem! While $[t]$ is indeed a step function, it is being subtracted by $f(t)$, which is also a step function. When you combine them, the jumps in $[t]$ and $f(t)$ perfectly cancel each other out in the neighborhood of $t=0$. It's a mathematical 'magic trick' that simplifies the entire problem.

Question

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

Options:

(A) 1

(B) $\sin(1)$

(C) -1

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you see a composite limit like $\lim_{x 	o 1} g(h(x-1))$, your first instinct might be panic. Think of this like an onion: we need to peel the outer layers to reach the core.

The core here is the inner function $h(x-1)$.

### The Power of Substitution

The expression $x-1$ inside the function is a classic hint to simplify the domain. Let's define a new variable $t = x - 1$.

As $x 	o 1$, $t$ naturally approaches $0$. This simple substitution transforms our intimidating limit into:

$$\lim_{t 	o 0} g(h(t))$$

Now, we are no longer dealing with $x$; we are dealing with the behavior of $h(t)$ as $t$ hovers around zero.

### Dissecting the Inner Beast

We need to understand $h(t) = 2[t] - f(t)$ near $t=0$. Let's look at the right side first, where $t \in (0, 1)$.

In this interval, the greatest integer function $[t]$ is $0$. For any positive $t$, our function $f(t)$ is defined as:

$$f(t) = \frac{t}{|t|} = \frac{t}{t} = 1$$

So, $h(t) = 2(0) - 1 = -1$.

Now, let's look at the left side, where $t \in (-1, 0)$. Here, $[t] = -1$.

For any negative $t$, $f(t)$ is defined as:

$$f(t) = \frac{t}{|t|} = \frac{t}{-t} = -1$$

Plugging these into our expression, we get:

$$h(t) = 2(-1) - (-1) = -2 + 1 = -1$$

Even at exactly $t=0$, where $[0] = 0$ and $f(0) = 1$, we get $h(0) = 2(0) - 1 = -1$.

### The 'Aha!' Moment

Whether we approach from the left or the right, or even sit exactly at zero, $h(t)$ is constantly $-1$. This is a beautiful, elegant result.

Because $h(t)$ is constant in the neighborhood of $0$, we don't need to worry about the limit of $g$ as it approaches $-1$; we simply need to evaluate the function $g$ at $-1$.

### The Final Evaluation

We are left with $g(-1)$. Looking at the definition of $g(x)$, it is a piecewise function:

$$g(x) = 
\begin{cases} 
\frac{\sin(x+1)}{x+1} & x 
eq -1 \
1 & x = -1 
\end{cases}$$

Since we need the value at $x = -1$, we look at the second condition. It explicitly tells us that $g(-1) = 1$.

And there you have it! By methodically breaking down the composite function, we found that the entire expression simplifies to **$1$**.

Select Answer:

Visualized Solution

The Composite Limit Challenge

Decoding $f (x)$ for Positive $x$

Decoding $f (x)$ for Negative $x$ and Zero

Substitution for Simplicity

Evaluating $h (t)$ on the Right of Zero

Evaluating $h (t)$ on the Left of Zero

Evaluating $h (t)$ exactly at Zero

The Constant Nature of the Inner Function

Evaluating $g (- 1)$

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Power of Substitution

Dissecting the Inner Beast

The 'Aha!' Moment

The Final Evaluation

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Select Answer:

Visualized Solution

The Composite Limit Challenge

Decoding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for Positive .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Decoding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for Negative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and Zero

Substitution for Simplicity

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(t) on the Right of Zero

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(t) on the Left of Zero

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(t) exactly at Zero