Let f(x) = { x - 1, x is even; 2x, x is odd }, x ∈ N. If for some a ∈ N, f(f(f(a))) = 21, then lim x

Q: Wait, why do we reject the case where $a = 23/4$?

Great question! The problem explicitly states that $a \in \mathbb{N}$, meaning $a$ must be a natural number (a positive integer). Since $23/4 = 5.75$, which is not an integer, this solution is physically impossible within the domain of the function.

Q: Why is $\left[ \frac{x}{12} \right] = 0$ as $x \to 12^-$?

The greatest integer function $[t]$ returns the largest integer less than or equal to $t$. As $x$ approaches $12$ from the left, $x$ is slightly less than $12$ (e.g., $11.99$). Therefore, $\frac{x}{12}$ is slightly less than $1$ (e.g., $0.99$). The largest integer less than or equal to $0.99$ is $0$.

Q: Can we just plug in $x=12$ for the whole limit expression?

Not quite! While you can plug in $x=12$ for the continuous part $\frac{|x|^3}{12}$, you cannot do that for the greatest integer function $\left[ \frac{x}{12} \right]$. The greatest integer function is discontinuous at every integer. You must evaluate the limit by considering the behavior of the function as $x$ approaches the target value from the left.

Question

Let $f(x) = \begin{cases} x - 1, & x 	ext{ is even}, \ 2x, & x 	ext{ is odd}, \end{cases} x \in \mathbb{N}$. If for some $a \in \mathbb{N}, f(f(f(a))) = 21$, then $\lim_{x 	o a^-} \left\{ \frac{|x|^3}{a} - \left[ \frac{x}{a} ight] ight\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :

Options:

(A) 121

(B) 144

(C) 169

(D) 225

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the piecewise function:

$$f(x) = \begin{cases} x - 1, & x 	ext{ is even} \ 2x, & x 	ext{ is odd} \end{cases}$$

Our objective is to find a natural number $a$ such that $f(f(f(a))) = 21$. We will evaluate this by considering the parity of $a$.

### Case 1: Assume $a$ is even

If $a$ is even, the first application yields $f(a) = a - 1$. Since $a$ is even, $a - 1$ is necessarily odd.

Applying the function a second time, we use the odd branch:

$$f(f(a)) = f(a - 1) = 2(a - 1) = 2a - 2$$

Note that $2a - 2$ is always even. Applying the function a third time using the even branch:

$$f(f(f(a))) = f(2a - 2) = (2a - 2) - 1 = 2a - 3$$

Setting this equal to $21$:

$$2a - 3 = 21 \Rightarrow 2a = 24 \Rightarrow a = 12$$

Since $12$ is an even natural number, this is a **valid solution**.

### Case 2: Assume $a$ is odd

If $a$ is odd, the first application yields $f(a) = 2a$, which is even. Applying the function a second time:

$$f(f(a)) = f(2a) = 2a - 1$$

Since $2a$ is even, $2a - 1$ is odd. Applying the function a third time using the odd branch:

$$f(f(f(a))) = f(2a - 1) = 2(2a - 1) = 4a - 2$$

Setting this equal to $21$:

$$4a - 2 = 21 \Rightarrow 4a = 23 \Rightarrow a = \frac{23}{4}$$

Because $\frac{23}{4}$ is not a natural number, we reject this case. Thus, the only valid value is **$a = 12$**.

### The Limit's Trap: Evaluating the Final Expression

We now evaluate the limit using our anchor value $a = 12$:

$$\lim_{x 	o 12^-} \left( \frac{|x|^3}{12} - \left[ \frac{x}{12} ight] ight)$$

As $x 	o 12^-$, $x$ approaches $12$ from the left, meaning $x < 12$. Consequently, the ratio $\frac{x}{12}$ is slightly less than $1$.

The greatest integer function $\left[ \frac{x}{12} ight]$ for any value in the interval $(0, 1)$ is $0$. Therefore:

$$\lim_{x 	o 12^-} \left[ \frac{x}{12} ight] = 0$$

For the first term, since $x$ is positive near $12$, we have $|x| = x$. Substituting $x = 12$:

$$\lim_{x 	o 12^-} \frac{|x|^3}{12} = \frac{12^3}{12} = 12^2 = 144$$

Combining these results, the final value is:

$$144 - 0 = \mathbf{144}$$

Select Answer:

Visualized Solution

Understanding the Piecewise Function

Case 1: Assuming $a$ is Even

Finding $f (f (a))$ for Case 1

Finding $f (f (f (a)))$ for Case 1

Solving for $a$ in Case 1

Case 2: Assuming $a$ is Odd

Finding $f (f (a))$ for Case 2

Finding $f (f (f (a)))$ for Case 2

Solving for $a$ in Case 2

Setting up the Limit

Evaluating the Greatest Integer Function

Evaluating the Cubic Term and Final Answer

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

Analyzing the Setup

Case 1

Case 2

The Limit's Trap

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Select Answer:

Visualized Solution

Understanding the Piecewise Function

Case 1: Assuming .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a is Even

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

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Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(f(f(a))) for Case 2

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