Questions: [ lim x rightarrow 3^+left(frac1x^2-9-frac1x-3right) ] (a) Describe the type of indeterminate form (if any) that is obtained by direct substitution. - (frac00) - (fracinftyinfty) - (0 cdot infty) - (1^infty) - (0^0) - (infty-infty) - (infty^0) - not indeterminate (b) Evaluate the limit, using L'Hôpital's Rule if necessary. Use a graphing utility to graph the function and verify the result. (If the limit is infinite or the limit does not otherwise exist, enter DNE.)

Solution Steps

To solve the given limit problem, we first need to identify the type of indeterminate form by substituting \( x = 3 \) into the expression. Then, we can use L'Hôpital's Rule to evaluate the limit if it results in an indeterminate form. Finally, we can graph the function to verify the result.

First, we need to determine the type of indeterminate form by direct substitution of \( x = 3 \) into the expression:

\[ \lim _{x \rightarrow 3^{+}}\left(\frac{1}{x^{2}-9}-\frac{1}{x-3}\right) \]

Substitute \( x = 3 \):

\[ \frac{1}{3^2 - 9} - \frac{1}{3 - 3} = \frac{1}{9 - 9} - \frac{1}{0} = \frac{1}{0} - \frac{1}{0} \]

This results in the form \(\infty - \infty\), which is an indeterminate form.

To resolve the indeterminate form, we simplify the expression:

\[ \frac{1}{x^2 - 9} - \frac{1}{x - 3} = \frac{1}{(x-3)(x+3)} - \frac{1}{x-3} \]

Combine the fractions over a common denominator:

\[ = \frac{1 - (x+3)}{(x-3)(x+3)} = \frac{1 - x - 3}{(x-3)(x+3)} = \frac{-x - 2}{(x-3)(x+3)} \]

\[ = \frac{-(x+2)}{(x-3)(x+3)} \]

Now, evaluate the limit as \( x \to 3^+ \):

\[ \lim_{x \to 3^+} \frac{-(x+2)}{(x-3)(x+3)} \]

Substitute \( x = 3 \):

\[ = \frac{-(3+2)}{(3-3)(3+3)} = \frac{-5}{0 \cdot 6} = \frac{-5}{0} \]

Since the denominator approaches zero from the positive side as \( x \to 3^+ \), the limit is \(-\infty\).

Final Answer