Let [t] denote the greatest integer ≤ t and {t} denote the fractional part of t. Then integral value

Q: Wait, why is $[1+x] = 0$ when $x \to 0^-$? Shouldn't it be $1$?

Great question! Remember, $x$ is slightly *less* than $0$. So $1+x$ is slightly less than $1$ (like $0.999...$). The greatest integer of any number just below $1$ is $0$. Always check the value inside the bracket carefully!

Q: Why can't we just use L'Hopital's rule here?

L'Hopital's rule is for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Here, we are dealing with step functions ($[x]$ and $\{x\}$). These functions are discontinuous and piecewise constant. We must evaluate them by their definition in the specific neighborhood of the limit, not by differentiation.

Q: I got $\alpha = 3$ and $\alpha = 1/3$. Why do we reject $1/3$?

The problem explicitly asks for the *integral* value of $\alpha$. In competitive exams, always re-read the question for constraints like 'integer', 'positive', or 'real'. Since $1/3$ is not an integer, it is mathematically valid but physically/contextually rejected.

Question

Let $[t]$ denote the greatest integer $\le t$ and $\{t\}$ denote the fractional part of $t$. Then integral value of $\alpha$ for which the left hand limit of the function $f(x) = [1 + x] + \frac{\alpha^{2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}$ at $x = 0$ is equal to $\alpha - \frac{4}{3}$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the function $f(x) = [1 + x] + \frac{\alpha^{2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}$. Our mission is to find the integral value of $\alpha$ such that the left-hand limit (LHL) at $x = 0$ equals $\alpha - \frac{4}{3}$.

To begin, we must define the behavior of $x$ as $x 	o 0^-$. This implies $x$ approaches zero from the negative side, specifically within the interval $(-1, 0)$.

### Phase 1: The Left-Hand Approach

When $x \in (-1, 0)$, the greatest integer function $[x]$ is constant. Specifically, $[x] = -1$.

Next, we evaluate the fractional part $\{x\}$. By definition, $\{x\} = x - [x]$. Substituting our value for $[x]$, we get $\{x\} = x - (-1) = x + 1$. As $x 	o 0^-$, the value of $\{x\}$ approaches $0 + 1 = 1$.

Finally, consider the term $[1 + x]$. Since $x$ is a small negative number, $1 + x$ is a value slightly less than $1$ (e.g., $0.999$). Therefore, $[1 + x] = 0$.

### Phase 2: The Algebraic Dance

Now, we substitute these values into the expression for $f(x)$. The denominator becomes:

$$2[x] + \{x\} = 2(-1) + 1 = -1$$

The numerator becomes:

$$\alpha^{2[x] + \{x\}} + [x] - 1 = \alpha^{-1} + (-1) - 1 = \alpha^{-1} - 2$$

Combining these, the function simplifies as follows:

$$f(x) = 0 + \frac{\alpha^{-1} - 2}{-1} = 2 - \alpha^{-1} = 2 - \frac{1}{\alpha}$$

### Phase 3: The Final Resolution

We are given that the LHL equals $\alpha - \frac{4}{3}$. We set up the following equation:

$$2 - \frac{1}{\alpha} = \alpha - \frac{4}{3}$$

Rearranging the terms to group the variables, we obtain:

$$\alpha + \frac{1}{\alpha} = 2 + \frac{4}{3} = \frac{10}{3}$$

By inspection of the equation $\alpha + \frac{1}{\alpha} = 3 + \frac{1}{3}$, we identify two possible values: $\alpha = 3$ or $\alpha = \frac{1}{3}$.

Since the problem explicitly requires $\alpha$ to be an **integral value**, we discard $\frac{1}{3}$. Thus, the final solution is **$\alpha = 3$**.

Enter Numerical Value:

Visualized Solution

Understanding the Function $f (x)$

Analyzing $x \to 0^{-}$

Evaluating $[x]$ for $x \to 0^{-}$

Evaluating ${x}$ for $x \to 0^{-}$

Evaluating $[1 + x]$ for $x \to 0^{-}$

Simplifying the Expression $2 [x] + {x}$

Calculating the LHL

Equating to Given Limit

Solving for $α$

Final Conclusion

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Enter Numerical Value:

Visualized Solution

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

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0−

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Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } {x} for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0−

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [1+x] for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0−

Simplifying the Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2[x]+{x}