Evaluate the following limits: x infinity ( x)/(sqrt(x)) and x 0 ( x)/(x)

As $x \to \infty$, $\ln x \to \infty$ and $\sqrt{x} \to \infty$. This is an indeterminate form of type $\frac{\infty}{\infty}$, so we can apply L'Hôpital's Rule.

Take the derivative of the numerator and the denominator:

Now, evaluate the limit of the ratio of these derivatives:

As $x \to \infty$, $\sqrt{x} \to \infty$, so $\frac{2}{\sqrt{x}} \to 0$.

The limit is $\boxed{0}$.

2) $\lim_{x \to 0} \frac{\tan x}{x}$

As $x \to 0$, $\tan x \to \tan 0 = 0$ and $x \to 0$. This is an indeterminate form of type $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

Take the derivative of the numerator and the denominator:

Now, evaluate the limit of the ratio of these derivatives:

Substitute $x=0$:

The limit is $\boxed{1}$.

3) $\lim_{x \to \infty} \frac{x^2}{e^x}$

As $x \to \infty$, $x^2 \to \infty$ and $e^x \to \infty$. This is an indeterminate form of type $\frac{\infty}{\infty}$, so we can apply L'Hôpital's Rule.

Take the derivative of the numerator and the denominator:

The new limit is $\lim_{x \to \infty} \frac{2x}{e^x}$. This is still an indeterminate form of type $\frac{\infty}{\infty}$.

Take the derivative of the new numerator and denominator:

The new limit is $\lim_{x \to \infty} \frac{2}{e^x}$.

As $x \to \infty$, $e^x \to \infty$, so $\frac{2}{e^x} \to 0$.

The limit is $\boxed{0}$.

4) $\lim_{x \to 0} \frac{\ln(1+x)}{x}$

As $x \to 0$, $\ln(1+x) \to \ln(1) = 0$ and $x \to 0$. This is an indeterminate form of type $\frac{0}{0}$, so we can apply L'Hôpital's Rule.

Take the derivative of the numerator and the denominator:

Now, evaluate the limit of the ratio of these derivatives:

Substitute $x=0$:

The limit is $\boxed{1}$.