Evaluate 3x3+4x2+7 / 6x3-5+9. Evaluate lim x5-32 / x-2. Evaluate lim x4-16 / x5-32.

here are the solutions to the problems.

Example 3: Evaluate $3x^3+4x^2+7$ $6x^3-5+9$

This question appears to be asking for the evaluation of a rational function, but it is missing the limit notation and the value $x$ approaches. Given the context of the other problems, we will assume it asks to evaluate the limit of the rational function as $x$ approaches infinity: $\lim_{x \to \infty} \frac{3x^3+4x^2+7}{6x^3-5+9}$

Step 1: Simplify the denominator. $\lim_{x \to \infty} \frac{3x^3+4x^2+7}{6x^3+4}$

Step 2: Divide every term in the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^3$. $\lim_{x \to \infty} \frac{\frac{3x^3}{x^3}+\frac{4x^2}{x^3}+\frac{7}{x^3}}{\frac{6x^3}{x^3}+\frac{4}{x^3}}$ $= \lim_{x \to \infty} \frac{3+\frac{4}{x}+\frac{7}{x^3}}{6+\frac{4}{x^3}}$

Step 3: Apply the limit. As $x \to \infty$, any term of the form $\frac{c}{x^n}$ approaches 0. $= \frac{3+0+0}{6+0}$ $= \frac{3}{6}$

Step 4: Simplify the fraction. $= \frac{1}{2}$ Assuming the question is $\lim_{x \to \infty} \frac{3x^3+4x^2+7}{6x^3-5+9}$, the value is $\boxed{\frac{1}{2}}$.

Question 4: Evaluate $\lim_{x \to 2} \frac{x^5-32}{x-2}$

When we substitute $x=2$ into the expression, we get $\frac{2^5-32}{2-2} = \frac{32-32}{0} = \frac{0}{0}$, which is an indeterminate form. We can use L'Hôpital's Rule.

Step 1: Differentiate the numerator and the denominator with respect to $x$. Derivative of the numerator $(x^5-32)$ is $5x^4$. Derivative of the denominator $(x-2)$ is $1$.

Step 2: Apply L'Hôpital's Rule by taking the limit of the ratio of the derivatives. $\lim_{x \to 2} \frac{5x^4}{1}$

Step 3: Substitute $x=2$ into the new expression. $5(2)^4$ $= 5(16)$ $= 80$ The limit is $\boxed{80}$.

Question 5: $\lim_{x \to 2} \frac{x^4-16}{x^5-32}$

When we substitute $x=2$ into the expression, we get $\frac{2^4-16}{2^5-32} = \frac{16-16}{32-32} = \frac{0}{0}$, which is an indeterminate form. We can use L'Hôpital's Rule.

Step 1: Differentiate the numerator and the denominator with respect to $x$. Derivative of the numerator $(x^4-16)$ is $4x^3$. Derivative of the denominator $(x^5-32)$ is $5x^4$.

Step 2: Apply L'Hôpital's Rule by taking the limit of the ratio of the derivatives. $\lim_{x \to 2} \frac{4x^3}{5x^4}$

Step 3: Simplify the expression. $\lim_{x \to 2} \frac{4}{5x}$

Step 4: Substitute $x=2$ into the simplified expression. $\frac{4}{5(2)}$ $= \frac{4}{10}$

Step 5: Simplify the fraction. $= \frac{2}{5}$ The limit is $\boxed{\frac{2}{5}}$.