The document contains a set of problems related to limits from a tutorial. I will solve a few of these limits step-by-step:

1. (a) $\lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h}$

To solve this, multiply both the numerator and denominator by the conjugate:

$\frac{\sqrt{4+h} - 2}{h} \cdot \frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2} = \frac{(4+h) - 4}{h(\sqrt{4+h} + 2)} = \frac{h}{h(\sqrt{4+h} + 2)} = \frac{1}{\sqrt{4+h} + 2}$

Now, take the limit as $h \to 0$:

$\lim_{h \to 0} \frac{1}{\sqrt{4+0} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$

So, the limit is $\frac{1}{4}$.

1. (b) $\lim_{h \to 0} \frac{h^2 - 2h}{h}$

Simplify the expression:

$\frac{h^2 - 2h}{h} = h - 2$

Now, take the limit as $h \to 0$:

$\lim_{h \to 0} (h - 2) = 0 - 2 = -2$

So, the limit is $-2$.

1. (c) $\lim_{h \to 0} \frac{(1-h^2) - (1+2h)}{3-h^2}$

Simplify the numerator:

$(1-h^2) - (1+2h) = 1 - h^2 - 1 - 2h = -h^2 - 2h$

The expression becomes:

$\lim_{h \to 0} \frac{-h^2 - 2h}{3 - h^2}$

As $h \to 0$, the numerator becomes 0 and the denominator becomes 3:

$\lim_{h \to 0} \frac{0}{3} = 0$

So, the limit is 0.

1. (d) $\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$

This is the definition of the derivative of $x^2$ at any point $x$. Expand the numerator:

$(x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$

Now, the expression becomes:

$\frac{2xh + h^2}{h} = 2x + h$

Take the limit as $h \to 0$:

$\lim_{h \to 0} (2x + h) = 2x + 0 = 2x$

So, the limit is $2x$.

Would you like further explanation or solutions for other parts?

Here are some related questions:

1. How do you apply the conjugate method for simplifying limits?
2. What are the key steps to identifying limits involving polynomials?
3. How do you interpret limits that define a derivative?
4. Why does multiplying by the conjugate help in evaluating limits involving square roots?
5. What does it mean if a limit is zero?

Tip: When solving limits, always check if a limit represents the derivative of a function at a point. It can greatly simplify the problem.