To find $\lim_{x \to 0^+} x^{\sqrt{x}}$, let's analyze the behavior of the function carefully.

Step 1: Rewrite the function

We can express $x^{\sqrt{x}}$ as an exponential: $x^{\sqrt{x}} = e^{\ln(x^{\sqrt{x}})} = e^{\sqrt{x} \ln(x)}.$

So, we now focus on evaluating the limit of the exponent $\sqrt{x} \ln(x)$ as $x \to 0^+$.

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Step 2: Analyze $\sqrt{x} \ln(x)$

Substitute $t = \sqrt{x}$, so $t^2 = x$. As $x \to 0^+$, we have $t \to 0^+$. Thus: $\sqrt{x} \ln(x) = t \ln(t^2).$ Using the property of logarithms, $\ln(t^2) = 2\ln(t)$, so: $t \ln(t^2) = t \cdot 2\ln(t) = 2t \ln(t).$

Now we analyze the behavior of $2t \ln(t)$ as $t \to 0^+$:

• $\ln(t) \to -\infty$ as $t \to 0^+$,
• $t \to 0^+$ approaches 0.

Thus, the term $t \ln(t)$ is an indeterminate form $0 \cdot (-\infty)$, and we use substitution or L'Hôpital's Rule to evaluate it.

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Step 3: Use substitution

Let $u = \ln(t)$, so $t = e^u$ and $u \to -\infty$ as $t \to 0^+$. Then: $t \ln(t) = e^u \cdot u.$ As $u \to -\infty$, $e^u \to 0$ faster than $u \to -\infty$. Thus: $e^u \cdot u \to 0.$

Therefore: $\sqrt{x} \ln(x) \to 0 \quad \text{as} \quad x \to 0^+.$

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Step 4: Conclude the result

Since $\sqrt{x} \ln(x) \to 0$, we have: $x^{\sqrt{x}} = e^{\sqrt{x} \ln(x)} \to e^0 = 1.$

Thus, the limit is: $\lim_{x \to 0^+} x^{\sqrt{x}} = 1.$

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Final Answer:

$\boxed{1}$

Would you like a deeper explanation of any step? Here are some related questions to expand your understanding:

1. What is the general behavior of $x^a$ as $x \to 0^+$ for different values of $a$?
2. How does L'Hôpital's Rule help in evaluating indeterminate forms like $0 \cdot (-\infty)$?
3. What is the significance of logarithmic transformations in limit problems?
4. Can $\lim_{x \to 0^-} x^{\sqrt{x}}$ be evaluated, and if so, what is the result?
5. How would the limit change if $x^{\sqrt{x}}$ was replaced by $x^{x^p}$ for some $p > 0$?

Tip: Transforming problems into exponential and logarithmic forms often simplifies limit evaluations significantly.