Use L'Hospital to determine the following limit. Use exact values. $$ \lim _{x \rightarrow 0^{+}} x \ln x=\square $$

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We start with the limit

$$
\lim_{x \rightarrow 0^{+}} x \ln x.
$$

This expression is an indeterminate form of type $$0 \cdot (-\infty)$$. To use L'Hôpital's Rule, we first rewrite it as a quotient. Notice that

$$
x \ln x = \frac{\ln x}{\frac{1}{x}}.
$$

As $$x \rightarrow 0^{+}$$, the numerator $$\ln x \rightarrow -\infty$$ and the denominator $$\frac{1}{x} \rightarrow \infty$$, giving a form $$\frac{-\infty}{\infty}$$. Now, we can apply L'Hôpital's Rule.

Differentiate the numerator and the denominator with respect to $$x$$:

- The derivative of $$\ln x$$ is $$\frac{1}{x}$$.
- The derivative of $$\frac{1}{x}$$ is $$-\frac{1}{x^2}$$.

Thus, the limit becomes

$$
\lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \rightarrow 0^{+}} \left( -x \right).
$$

Now, as $$x \rightarrow 0^{+}$$,

$$
-x \rightarrow 0.
$$

Therefore, the original limit is

$$
\lim_{x \rightarrow 0^{+}} x \ln x = 0.
$$
The limit is 0.

Use L’Hospital to determine the following limit. Use
exact values.

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Use L’Hospital to determine the following limit. Use exact values.