Find the domain of the following functions: 1. f(x) = x/(x2+4) 2. f(x) = sqrt(x−4) 3. f(x) = sqrt(x2−16) 4. f(x) = sqrt(3−x)

Here are the solutions to the problems:

Domain Problems

1. To determine the domain of $\frac{x}{x^2+4}$, we need to ensure the denominator is not zero.

   Step 1: Set the denominator equal to zero.

   $$x^2+4 = 0$$

   Step 2: Solve for $x$.

   $$x^2 = -4$$

   There are no real values of $x$ for which $x^2 = -4$. Thus, the denominator is never zero.
   The domain is all real numbers.

   $$\boxed{(-\infty, \infty)}$$

2. To find the domain of $\sqrt{x-4}$, the expression under the square root must be non-negative.

   Step 1: Set the expression under the square root greater than or equal to zero.

   $$x-4 \ge 0$$

   Step 2: Solve for $x$.

   $$x \ge 4$$

   The domain is all real numbers greater than or equal to 4.

   $$\boxed{[4, \infty)}$$

3. To determine the domain of $\sqrt{x^2-16}$, the expression under the square root must be non-negative.

   Step 1: Set the expression under the square root greater than or equal to zero.

   $$x^2-16 \ge 0$$

   Step 2: Factor the inequality.

   $$(x-4)(x+4) \ge 0$$

   Step 3: Find the critical points where the expression equals zero, which are $x=4$ and $x=-4$. Test intervals to find where the inequality holds true.
   The expression is non-negative when $x \le -4$ or $x \ge 4$.

   $$\boxed{(-\infty, -4] \cup [4, \infty)}$$

4. To find the domain of $\sqrt{3-x}$, the expression under the square root must be non-negative.

   Step 1: Set the expression under the square root greater than or equal to zero.

   $$3-x \ge 0$$

   Step 2: Solve for $x$.

   $$3 \ge x$$ $$x \le 3$$

   The domain is all real numbers less than or equal to 3.

   $$\boxed{(-\infty, 3]}$$

5. If $f(x) = \sqrt{4-x^2}$, to find the domain, the expression under the square root must be non-negative.

   Step 1: Set the expression under the square root greater than or equal to zero.

   $$4-x^2 \ge 0$$

   Step 2: Rearrange the inequality.

   $$x^2 \le 4$$

   Step 3: Take the square root of both sides, remembering to consider both positive and negative roots.

   $$-2 \le x \le 2$$

   The domain is all real numbers between $-2$ and 2, inclusive.

   $$\boxed{[-2, 2]}$$

6. To determine the domain and range of $|x|$:

   Step 1: For the domain, consider all possible real values for $x$. The absolute value function is defined for all real numbers.

   Step 2: For the range, consider the output values of the absolute value function. The absolute value of any real number is always non-negative.
   Domain: $\boxed{(-\infty, \infty)}$
   Range: $\boxed{[0, \infty)}$

Limits of a Function

1. Evaluate $\lim_{x \to 0} \frac{7^x-3}{5^x+3}$

   Step 1: Substitute $x=0$ into the expression.

   $$\lim_{x \to 0} \frac{7^x-3}{5^x+3} = \frac{7^0-3}{5^0+3}$$

   Step 2: Simplify the expression.

   $$= \frac{1-3}{1+3} = \frac{-2}{4} = -\frac{1}{2}$$ $$\boxed{-\frac{1}{2}}$$

2. Evaluate $\lim_{x \to \infty} \frac{x^3-x^2+4x}{2x^3+2x^2-7}$

   Step 1: Identify the highest power of $x$ in the numerator and denominator. Both are $x^3$.

   Step 2: The limit as $x \to \infty$ of a rational function where the degrees of the numerator and denominator are equal is the ratio of their leading coefficients.

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

3. Evaluate $\lim_{x \to \infty} \frac{x}{x-6}$

   Step 1: Identify the highest power of $x$ in the numerator and denominator. Both are $x^1$.

   Step 2: The limit as $x \to \infty$ of a rational function where the degrees of the numerator and denominator are equal is the ratio of their leading coefficients.

   $$\lim_{x \to \infty} \frac{x}{x-6} = \frac{1}{1} = 1$$ $$\boxed{1}$$

4. Compute $\lim_{x \to 6} \frac{x-6}{x^2-36}$

   Step 1: Direct substitution results in $\frac{0}{0}$, an indeterminate form. Factor the denominator.

   $$\lim_{x \to 6} \frac{x-6}{(x-6)(x+6)}$$

   Step 2: Cancel out the common factor $(x-6)$.

   $$= \lim_{x \to 6} \frac{1}{x+6}$$

   Step 3: Substitute $x=6$ into the simplified expression.

   $$= \frac{1}{6+6} = \frac{1}{12}$$ $$\boxed{\frac{1}{12}}$$

5. Evaluate $\lim_{x \to 6} \frac{x-6}{x^2+36}$

   Step 1: Substitute $x=6$ into the expression.

   $$\lim_{x \to 6} \frac{x-6}{x^2+36} = \frac{6-6}{6^2+36}$$

   Step 2: Simplify the expression.

   $$= \frac{0}{36+36} = \frac{0}{72} = 0$$ $$\boxed{0}$$

6. Evaluate $\lim_{x \to 1} \frac{x^2-2x+1}{x-1}$

   Step 1: Direct substitution results in $\frac{0}{0}$, an indeterminate form. Factor the numerator.

   $$\lim_{x \to 1} \frac{(x-1)^2}{x-1}$$

   Step 2: Cancel out the common factor $(x-1)$.

   $$= \lim_{x \to 1} (x-1)$$

   Step 3: Substitute $x=1$ into the simplified expression.

   $$= 1-1 = 0$$ $$\boxed{0}$$

7. Evaluate $\lim_{x \to 2} \frac{x-2}{x^2+2x-6}$

   Step 1: Substitute $x=2$ into the expression.

   $$\lim_{x \to 2} \frac{x-2}{x^2+2x-6} = \frac{2-2}{2^2+2(2)-6}$$

   Step 2: Simplify the expression.

   $$= \frac{0}{4+4-6} = \frac{0}{2} = 0$$ $$\boxed{0}$$

8. Evaluate $\lim_{x \to 4} \frac{x^2-16}{x-4}$

   Step 1: Direct substitution results in $\frac{0}{0}$, an indeterminate form. Factor the numerator.

   $$\lim_{x \to 4} \frac{(x-4)(x+4)}{x-4}$$

   Step 2: Cancel out the common factor $(x-4)$.

   $$$$