HE 4.3. Evaluate (1) lim sin πx/(x-1), (2) lim x tan(1/x), (3) lim (2+5x+x sin x)/(x2-3x), (4) lim sqrt(x2+3x)-x.

As $x \to 1$, the numerator $\sin(\pi \cdot 1) = \sin \pi = 0$. The denominator $1-1 = 0$. This is an indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule.

Take the derivative of the numerator and the denominator:

The limit becomes:

Substitute $x=1$ into the expression:

The value of the limit is $\boxed{-\pi}$.

2) $\lim_{x \to \infty} x \tan(1/x)$

Let $y = \frac{1}{x}$. As $x \to \infty$, $y \to 0$.
The expression becomes:

This is an indeterminate form $\frac{0}{0}$ as $\tan 0 = 0$.

Take the derivative of the numerator and the denominator with respect to $y$:

The limit becomes:

Substitute $y=0$ into the expression:

The value of the limit is $\boxed{1}$.

3) $\lim_{x \to \infty} \frac{2 + 5x + x \sin x}{x^2 - 3x}$

For $\lim_{x \to \infty} \frac{\sin x}{x}$, we know that $-1 \le \sin x \le 1$. Dividing by $x$ (for $x>0$):

By the Squeeze Theorem, since $\lim_{x \to \infty} -\frac{1}{x} = 0$ and $\lim_{x \to \infty} \frac{1}{x} = 0$, then $\lim_{x \to \infty} \frac{\sin x}{x} = 0$.

The value of the limit is $\boxed{0}$.

4) $\lim_{x \to \infty} \sqrt{x^2 + 3x} - x$

As $x \to \infty$, this is an $\infty - \infty$ indeterminate form.

For $x \to \infty$, $\sqrt{x^2 + 3x} = \sqrt{x^2(1 + \frac{3}{x})} = |x|\sqrt{1 + \frac{3}{x}}$. Since $x \to \infty$, $x > 0$, so $|x|=x$.

As $x \to \infty$, $\frac{3}{x} \to 0$.

The value of the limit is $\boxed{\frac{3}{2}}$.