If `a in I`, then value of a for which `lim_(xrarra) (tan([x^(2)]-[x]^(3)))/((x-a)^(3))` exists finitely, is /are

Options

A. 0
B. 1
C. `-1`
D. 2

Explanation

To solve the limit problem, we need to find the values of \( a \) for which the limit

\[ \lim_{x \to a} \frac{\tan(\lfloor x^3 \rfloor - \lfloor x \rfloor^3)}{(x - a)^3} \]

exists finitely, where \( a \) is an integer.

### Step 1: Define the function

Let

\[ f(x) = \tan(\lfloor x^3 \rfloor - \lfloor x \rfloor^3) \]

We need to evaluate \( f(a) \) and the left-hand and right-hand limits as \( x \) approaches \( a \).

### Step 2: Calculate \( f(a) \)

Substituting \( x = a \):

\[ f(a) = \tan(\lfloor a^3 \rfloor - \lfloor a \rfloor^3) \]

Since \( a \) is an integer, we have:

\[ \lfloor a^3 \rfloor = a^3 \quad \text{and} \quad \lfloor a \rfloor = a \]

Thus,

\[ f(a) = \tan(a^3 - a^3) = \tan(0) = 0 \]

### Step 3: Calculate the right-hand limit (RHL)

For the right-hand limit as \( x \to a^+ \):

\[ \lim_{x \to a^+} f(x) = \lim_{x \to a^+} \tan(\lfloor x^3 \rfloor - \lfloor x \rfloor^3) \]

As \( x \) approaches \( a \) from the right, \( \lfloor x^3 \rfloor = a^3 \) and \( \lfloor x \rfloor = a \). Therefore:

\[ \lim_{x \to a^+} f(x) = \tan(a^3 - a^3) = \tan(0) = 0 \]

### Step 4: Calculate the left-hand limit (LHL)

For the left-hand limit as \( x \to a^- \):

\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^-} \tan(\lfloor x^3 \rfloor - \lfloor x \rfloor^3) \]

As \( x \) approaches \( a \) from the left, \( \lfloor x^3 \rfloor = a^3 - 1 \) (since \( x^3 < a^3 \)) and \( \lfloor x \rfloor = a - 1 \). Thus:

\[ \lim_{x \to a^-} f(x) = \tan((a^3 - 1) - (a - 1)^3) \]

Expanding \( (a - 1)^3 \):

\[ (a - 1)^3 = a^3 - 3a^2 + 3a - 1 \]

So,

\[ \lim_{x \to a^-} f(x) = \tan((a^3 - 1) - (a^3 - 3a^2 + 3a - 1)) = \tan(3a^2 - 3a) \]

### Step 5: Set the limits equal for existence

For the limit to exist, we need:

\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) = 0 \]

This means:

\[ \tan(3a(a - 1)) = 0 \]

The tangent function is zero when its argument is a multiple of \( \pi \):

\[ 3a(a - 1) = k\pi \quad \text{for some integer } k \]

### Step 6: Solve for \( a \)

The simplest case is when \( 3a(a - 1) = 0 \):

1. \( a = 0 \)
2. \( a - 1 = 0 \Rightarrow a = 1 \)

Thus, the values of \( a \) for which the limit exists finitely are:

\[ a = 0 \quad \text{and} \quad a = 1 \]

### Final Answer

The values of \( a \) for which the limit exists finitely are \( a = 0 \) and \( a = 1 \).

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