The value of `lim_(xrarr0^(+))(min(t^2-6(t+x)+12)+[(log_(e)(1+2x))/(x)])` , where , `t in R` , [y] denotes greatest integer less than or equal to y, is :

Options

A. 5
B. 3
C. 6
D. 9

Answer

To solve the limit problem, we need to evaluate the following expression:

\[ \lim_{x \to 0^+} \left( \min(t^2 - 6(t + x) + 12) + \left[\frac{\log_e(1 + 2x)}{x}\right] \right) \]

### Step 1: Simplify the first part of the limit

First, we simplify the expression inside the minimum function:

\[ t^2 - 6(t + x) + 12 = t^2 - 6t - 6x + 12 \]

This can be rearranged as:

\[ t^2 - 6t + 12 - 6x \]

### Step 2: Evaluate the minimum function

Now, we need to find the minimum of \( t^2 - 6t + 12 - 6x \). The expression \( t^2 - 6t + 12 \) can be rewritten as:

\[ (t - 3)^2 + 3 \]

Thus, we have:

\[ \min((t - 3)^2 + 3 - 6x) \]

As \( x \to 0^+ \), the term \( -6x \) approaches \( 0 \). Therefore, we need to evaluate:

\[ \min((t - 3)^2 + 3) \]

The minimum value of \( (t - 3)^2 \) occurs when \( t = 3 \), giving us:

\[ \min((t - 3)^2 + 3) = 0 + 3 = 3 \]

### Step 3: Evaluate the second part of the limit

Next, we evaluate the second part of the limit:

\[ \frac{\log_e(1 + 2x)}{x} \]

Using L'Hôpital's Rule, since both the numerator and denominator approach \( 0 \) as \( x \to 0^+ \):

1. Differentiate the numerator: \( \frac{d}{dx}[\log_e(1 + 2x)] = \frac{2}{1 + 2x} \)
2. Differentiate the denominator: \( \frac{d}{dx}[x] = 1 \)

Applying L'Hôpital's Rule:

\[ \lim_{x \to 0^+} \frac{\log_e(1 + 2x)}{x} = \lim_{x \to 0^+} \frac{2}{1 + 2x} = \frac{2}{1} = 2 \]

### Step 4: Combine the results

Now, we combine the results of both parts of the limit:

\[ \lim_{x \to 0^+} \left( \min(t^2 - 6(t + x) + 12) + \left[\frac{\log_e(1 + 2x)}{x}\right] \right) = 3 + 2 = 5 \]

### Final Result

Thus, the value of the limit is:

\[ \boxed{5} \]