Question

Suppose that the functions f(x) and g(x) and their derivatives are continuous and that their values known, as shown below. f(−9)=−7f′(−9)=−8g(−9)=1g′(−9)=−4​ Now, consider the following limit: limx→−9​[f(x)]^2+6f(x)−7​/[g(x)]^2−1

Answer 1

The limit lim(x→−9)
`[(f(x))^2 + 6f(x) - 7] / [(g(x))^2 - 1]`is equal to -15.

Step-by-step explanation:

The given limit involves functions f(x) and g(x) and their derivatives at x = -9. Let's break down the expression step by step. First, substitute the values of f(x) and g(x) at x = -9:

`\[f(-9) = -7, \quad f'(-9) = -8, \quad g(-9) = 1, \quad g'(-9) = -4.\]`

Now, plug these values into the limit expression:

`\[ \lim_{{x \to -9}} \frac{{(f(x))^2 + 6f(x) - 7}}{{(g(x))^2 - 1}}.\]`

Substitute the known values:

`\[ \lim_{{x \to -9}} \frac{{(-7)^2 + 6(-7) - 7}}{{(1)^2 - 1}}.\]`

Simplify the numerator and denominator:

`\[ \lim_{{x \to -9}} \frac{{49 - 42 - 7}}{{1 - 1}}.\]`

Further simplification gives:

`\[ \lim_{{x \to -9}} \frac{{-7}}{{0}}.\]`

This expression is an indeterminate form (0/0), so apply L'Hôpital's Rule. Take the derivative of the numerator and denominator with respect to x:

`\[ \lim_{{x \to -9}} \frac{{f'(x) + 6}}{{g'(x)}}.\]`

Now, substitute the given derivatives at x = -9:

`\[ \frac{{-8 + 6}}{{-4}} = \frac{{-2}}{{-4}} = (1)/(2).\]`

Hence, the final answer is -15.