The problem asks us to determine the limits for four different functions. a) (i) $\lim_{x \to 3} (x^4 + 4x^2 - 3)$ a) (ii) $\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8-z}$ b) Given $\lim_{x \to 0} f(x) = 6$, $\lim_{x \to 0} g(x) = -4$, and $\lim_{x \to 0} h(x) = -1$, compute $\lim_{x \to 0} \sqrt{\frac{f(x)}{h(x) - g(x)}}$ c) For $f(x) = \frac{3x^7 - 4x^2 + 1}{5 - 10x^2}$, evaluate $\lim_{x \to \infty} f(x)$ d) Find $\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2}}{x-2}$

1. Problem Description

The problem asks us to determine the limits for four different functions.

a) (i) $\lim_{x \to 3} (x^4 + 4x^2 - 3)$

a) (ii) $\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8-z}$

b) Given $\lim_{x \to 0} f(x) = 6$, $\lim_{x \to 0} g(x) = -4$, and $\lim_{x \to 0} h(x) = -1$, compute $\lim_{x \to 0} \sqrt{\frac{f(x)}{h(x) - g(x)}}$

c) For $f(x) = \frac{3x^7 - 4x^2 + 1}{5 - 10x^2}$, evaluate $\lim_{x \to \infty} f(x)$

d) Find $\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2}}{x-2}$

2. Solution Steps

a) (i) $\lim_{x \to 3} (x^4 + 4x^2 - 3)$

Since this is a polynomial, we can directly substitute $x = 3$ into the expression.

$\lim_{x \to 3} (x^4 + 4x^2 - 3) = (3^4 + 4(3^2) - 3) = 81 + 4(9) - 3 = 81 + 36 - 3 = 114$

a) (ii) $\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8-z}$

We first try to substitute $z=8$ directly into the expression:

$\frac{2(8^2) - 17(8) + 8}{8-8} = \frac{128 - 136 + 8}{0} = \frac{0}{0}$.

Since we get an indeterminate form, we factorize the numerator.

$2z^2 - 17z + 8 = (2z-1)(z-8) = -(2z-1)(8-z)$.

Then

$\lim_{z \to 8} \frac{2z^2 - 17z + 8}{8-z} = \lim_{z \to 8} \frac{-(2z-1)(8-z)}{8-z} = \lim_{z \to 8} -(2z-1) = -(2(8)-1) = -(16-1) = -15$

b) $\lim_{x \to 0} \sqrt{\frac{f(x)}{h(x) - g(x)}}$

Since we are given the limits of $f(x)$, $g(x)$, and $h(x)$, we can use limit properties.

$\lim_{x \to 0} \sqrt{\frac{f(x)}{h(x) - g(x)}} = \sqrt{\frac{\lim_{x \to 0} f(x)}{\lim_{x \to 0} h(x) - \lim_{x \to 0} g(x)}} = \sqrt{\frac{6}{-1 - (-4)}} = \sqrt{\frac{6}{-1 + 4}} = \sqrt{\frac{6}{3}} = \sqrt{2}$

c) $\lim_{x \to \infty} \frac{3x^7 - 4x^2 + 1}{5 - 10x^2}$

We divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$.

$\lim_{x \to \infty} \frac{3x^7 - 4x^2 + 1}{5 - 10x^2} = \lim_{x \to \infty} \frac{\frac{3x^7}{x^2} - \frac{4x^2}{x^2} + \frac{1}{x^2}}{\frac{5}{x^2} - \frac{10x^2}{x^2}} = \lim_{x \to \infty} \frac{3x^5 - 4 + \frac{1}{x^2}}{\frac{5}{x^2} - 10}$

As $x \to \infty$, $3x^5 \to \infty$ and $\frac{1}{x^2} \to 0$ and $\frac{5}{x^2} \to 0$.

So the limit is $\lim_{x \to \infty} \frac{3x^5 - 4 + \frac{1}{x^2}}{\frac{5}{x^2} - 10} = \frac{\infty}{-10} = -\infty$

d) $\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2}}{x-2}$

If we substitute $x=2$, we get $\frac{0}{0}$. We multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x} + \sqrt{2}$.

$\lim_{x \to 2} \frac{\sqrt{x} - \sqrt{2}}{x-2} = \lim_{x \to 2} \frac{(\sqrt{x} - \sqrt{2})(\sqrt{x} + \sqrt{2})}{(x-2)(\sqrt{x} + \sqrt{2})} = \lim_{x \to 2} \frac{x - 2}{(x-2)(\sqrt{x} + \sqrt{2})} = \lim_{x \to 2} \frac{1}{\sqrt{x} + \sqrt{2}}$

Now substitute $x=2$.

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

3. Final Answer

a) (i) 114

a) (ii) -15

b) $\sqrt{2}$

c) $-\infty$

d) $\frac{\sqrt{2}}{4}$