https://file.aiquickdraw.com/mathbot/file/1730159759638rbtz7vxd.jpg

Let’s analyze and solve each of the limits provided.

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### Problem 1: 
\[
\lim_{x \to 0^+} \sqrt{x} \ln x
\]

#### Solution:
Rewrite \(\sqrt{x}\) as \(x^{1/2}\):
\[
\lim_{x \to 0^+} x^{1/2} \ln x
\]

Now we have a form of \(0 \cdot -\infty\), which is an indeterminate form. To use L'Hospital's Rule, rewrite it as:
\[
\lim_{x \to 0^+} \frac{\ln x}{x^{-1/2}}
\]

Now, as \(x \to 0^+\), \(\ln x \to -\infty\) and \(x^{-1/2} \to \infty\), giving an indeterminate form \(\frac{-\infty}{\infty}\). Now apply L'Hospital's Rule by differentiating the numerator and denominator with respect to \(x\):

\[
= \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{2} x^{-3/2}} = \lim_{x \to 0^+} \frac{-2}{x^{1/2}} = -\infty
\]

So, the limit is:
\[
\boxed{-\infty}
\]

---

### Problem 2:
\[
\lim_{x \to \infty} x^2 e^x
\]

#### Solution:
As \(x \to \infty\), \(x^2 \to \infty\) and \(e^x \to \infty\), giving an indeterminate form \(\infty \cdot \infty\).

Since \(e^x\) grows faster than any polynomial term, this limit is straightforward:

\[
\lim_{x \to \infty} x^2 e^x = \infty
\]

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

---

### Problem 3:
\[
\lim_{x \to \infty} \left( \sqrt{x^2 + x} - x \right)
\]

#### Solution:
Rewrite the expression under the limit:
\[
\lim_{x \to \infty} \left( \sqrt{x^2 + x} - x \right)
\]

To simplify, multiply by the conjugate:
\[
= \lim_{x \to \infty} \frac{(\sqrt{x^2 + x} - x)(\sqrt{x^2 + x} + x)}{\sqrt{x^2 + x} + x}
\]

The numerator simplifies as follows:
\[
= \lim_{x \to \infty} \frac{x^2 + x - x^2}{\sqrt{x^2 + x} + x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + x} + x}
\]

Now factor out \(x\) from the square root in the denominator:
\[
= \lim_{x \to \infty} \frac{x}{x \sqrt{1 + \frac{1}{x}} + x} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}
\]

As \(x \to \infty\), \(\frac{1}{x} \to 0\), so we get:
\[
= \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}
\]

Therefore, the limit is:
\[
\boxed{\frac{1}{2}}
\]

---

Would you like further explanation on any of these steps, or do you have any specific questions?

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Here are five follow-up questions based on these solutions:
1. How can L'Hospital's Rule be applied to limits involving logarithmic and power functions?
2. What are some alternative methods to L'Hospital's Rule for finding limits?
3. Why does \(e^x\) grow faster than any polynomial term as \(x \to \infty\)?
4. How does multiplying by the conjugate help in simplifying limit problems?
5. How can we determine when a limit will approach infinity versus zero?

**Tip:** When dealing with limits at infinity, recognize the relative growth rates of different functions, such as exponential, polynomial, and logarithmic functions, to predict the limit behavior.

Find the limit: 1. lim(x→0+) √x ln(x), 2. lim(x→∞) x^2 e^x, 3. lim(x→∞) (√(x^2 + x) - x)

Explore solutions to challenging limit problems using techniques like L'Hospital's Rule and conjugates. This guide includes limits such as lim(x→0+) √x ln(x), lim(x→∞) x^2 e^x, and lim(x→∞) (√(x^2 + x) - x), with step-by-step explanations for understanding exponential growth and indeterminate forms.

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