Q4. Evaluate limy->1 (y2 - 1) / (y - 1). Q5. Find limx->inf (5x2 + 3x - 1) / (2x3 - 7). Q6. Evaluate limx->0 (1-cosx)/x2 * limx->0 (1-cosx)/x2. Q7. Determine lim_x->-2 (x2+5x+6)/(x+2).

Substituting $x=1$ gives $\frac{1^3 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form.

Here, $a=x$ and $b=1$, so $x^3 - 1 = (x-1)(x^2+x+1)$.

Now, substitute $x=1$:

The limit is $\boxed{3}$.

Q5. Find $\lim_{x \to \infty} \frac{5x^2 + 3x - 1}{2x^3 - 7}$

The highest power of $x$ in the denominator is $x^3$.

As $x \to \infty$, terms like $\frac{5}{x}$, $\frac{3}{x^2}$, $\frac{1}{x^3}$, and $\frac{7}{x^3}$ all approach 0.

The limit is $\boxed{0}$.

Q6. Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2} - \lim_{x \to 0} \left(\frac{1 - \cos x}{x^2}\right)$

The expression is in the form $A - A$, where $A = \lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Since any quantity subtracted from itself is zero, the result is 0.

The value is $\boxed{0}$.

Q7. Determine $\lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2}$

Substituting $x=-2$ gives $\frac{(-2)^2 + 5(-2) + 6}{-2 + 2} = \frac{4 - 10 + 6}{0} = \frac{0}{0}$, which is an indeterminate form.

The quadratic expression $x^2 + 5x + 6$ can be factored into $(x+2)(x+3)$.

Now, substitute $x=-2$:

The limit is $\boxed{1}$.