find all values of x where the function is discontinuous. for each value of x, give the limit of the…

find all values of x where the function is discontinuous. for each value of x, give the limit of the function at the value of x. be sure to note when the limit doesnt exist. r(x)=ln|2x/(x - 9)| select the correct choice below and, if necessary, fill in the answer box(es) within your choice. (use a comma to separate answers as needed.) oa. f is discontinuous at the single value x = . the limit is . ob. f is discontinuous at the single value x = . the limit does not exist and is not ∞ or -∞. oc. f is discontinuous at the two values x = . the limit for the smaller value is . the limit for the larger value is . od. f is discontinuous at the two values x = . the limit for the smaller value is . the limit for the larger value does not exist and is not ∞ or -∞. oe. f is discontinuous at the two values x = . the limit for the smaller value does not exist and is not ∞ or -∞. the limit for the larger value is . of. f is discontinuous over the interval . the limit is . (type your answer in interval notation.) og. f is discontinuous over the interval . the limit does not exist and is not ∞ or -∞. (type your answer in interval notation.)

find all values of x where the function is discontinuous. for each value of x, give the limit of the function at the value of x. be sure to note when the limit doesnt exist. r(x)=ln|2x/(x - 9)| select the correct choice below and, if necessary, fill in the answer box(es) within your choice. (use a comma to separate answers as needed.) oa. f is discontinuous at the single value x = . the limit is . ob. f is discontinuous at the single value x = . the limit does not exist and is not ∞ or -∞. oc. f is discontinuous at the two values x = . the limit for the smaller value is . the limit for the larger value is . od. f is discontinuous at the two values x = . the limit for the smaller value is . the limit for the larger value does not exist and is not ∞ or -∞. oe. f is discontinuous at the two values x = . the limit for the smaller value does not exist and is not ∞ or -∞. the limit for the larger value is . of. f is discontinuous over the interval . the limit is . (type your answer in interval notation.) og. f is discontinuous over the interval . the limit does not exist and is not ∞ or -∞. (type your answer in interval notation.)

Answer

Explanation:

Step1: Identify domain - issues

The natural - logarithm function (y = \ln(u)) is undefined when (u\leqslant0) and the rational function (\frac{2x}{x - 9}) is undefined when (x=9). First, find when (\frac{2x}{x - 9}\leqslant0) and (x = 9). Solve (\frac{2x}{x - 9}=0), we get (x = 0) (by setting the numerator equal to 0). The vertical asymptote of (y=\frac{2x}{x - 9}) is (x = 9) (by setting the denominator equal to 0).

Step2: Analyze the sign of (\frac{2x}{x - 9})

We consider the intervals ((-\infty,0)), ((0,9)) and ((9,\infty)). For (x\in(-\infty,0)), if (x=-1), then (\frac{2x}{x - 9}=\frac{-2}{-1 - 9}=\frac{-2}{-10}=\frac{1}{5}>0). For (x\in(0,9)), if (x = 1), then (\frac{2x}{x - 9}=\frac{2}{1 - 9}=-\frac{1}{4}<0). For (x\in(9,\infty)), if (x = 10), then (\frac{2x}{x - 9}=\frac{20}{10 - 9}=20>0). The function (r(x)=\ln\left|\frac{2x}{x - 9}\right|) is discontinuous when (\frac{2x}{x - 9}\leqslant0) or (x = 9). The values of (x) for which the function is discontinuous are (x = 0) and (x = 9).

Step3: Find the limit as (x\to0)

(\lim_{x\to0}\ln\left|\frac{2x}{x - 9}\right|=\ln\left|\frac{2\times0}{0 - 9}\right|=\ln(0)=-\infty).

Step4: Find the limit as (x\to9)

(\lim_{x\to9^{-}}\frac{2x}{x - 9}=-\infty) and (\lim_{x\to9^{+}}\frac{2x}{x - 9}=\infty), so (\lim_{x\to9}\ln\left|\frac{2x}{x - 9}\right|) does not exist.

Answer:

D. (f) is discontinuous at the two values (x = 0,9). The limit for the smaller value (x = 0) is (-\infty). The limit for the larger value (x = 9) does not exist and is not (\infty) or (-\infty).