The value of `lim_(xrarr0^(+)){x^(x^(2))+x^((x^(x)))}` is equal to

Options

A. 0
B. 1
C. 2
D. `(1)/(2)`

Answer

To solve the limit \( \lim_{x \to 0^+} \left( x^{x^2} + x^{x^x} \right) \), we will break it down into two parts and evaluate each limit separately.

### Step 1: Evaluate \( \lim_{x \to 0^+} x^{x^2} \)

We can rewrite \( x^{x^2} \) using the exponential function:
\[ x^{x^2} = e^{x^2 \ln x} \]
Now we need to evaluate the limit:
\[ \lim_{x \to 0^+} x^2 \ln x \]
As \( x \to 0^+ \), \( \ln x \to -\infty \) and \( x^2 \to 0 \). This creates an indeterminate form \( 0 \cdot (-\infty) \). We can rewrite it as:
\[ \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x^2}} \]
Now we can apply L'Hôpital's Rule since we have the form \( \frac{-\infty}{\infty} \).

### Step 2: Apply L'Hôpital's Rule

Taking the derivative of the numerator and the denominator:
- Derivative of \( \ln x \) is \( \frac{1}{x} \)
- Derivative of \( \frac{1}{x^2} \) is \( -\frac{2}{x^3} \)

So we have:
\[ \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{2}{x^3}} = \lim_{x \to 0^+} -\frac{x^3}{2x} = \lim_{x \to 0^+} -\frac{x^2}{2} = 0 \]
Thus,
\[ \lim_{x \to 0^+} x^2 \ln x = 0 \]
Therefore,
\[ \lim_{x \to 0^+} x^{x^2} = e^0 = 1 \]

### Step 3: Evaluate \( \lim_{x \to 0^+} x^{x^x} \)

Next, we evaluate \( x^{x^x} \):
\[ x^{x^x} = e^{x^x \ln x} \]
Now we need to evaluate \( \lim_{x \to 0^+} x^x \ln x \). First, we find \( \lim_{x \to 0^+} x^x \):
\[ x^x = e^{x \ln x} \]
Again, we need to evaluate:
\[ \lim_{x \to 0^+} x \ln x \]
This is also an indeterminate form \( 0 \cdot (-\infty) \), so we rewrite it:
\[ \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} \]
Applying L'Hôpital's Rule:
- Derivative of \( \ln x \) is \( \frac{1}{x} \)
- Derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \)

Thus,
\[ \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} -x = 0 \]
So,
\[ \lim_{x \to 0^+} x \ln x = 0 \implies \lim_{x \to 0^+} x^x = e^0 = 1 \]
Therefore,
\[ \lim_{x \to 0^+} x^{x^x} = e^{1 \cdot \ln x} = e^{\ln x} = x \]
As \( x \to 0^+ \), \( x \to 0 \).

### Step 4: Combine the Results

Now we combine both limits:
\[ \lim_{x \to 0^+} \left( x^{x^2} + x^{x^x} \right) = 1 + 0 = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]