consider the following function.\n\n$f(x)=\\frac{x - 4}{x^{2}-16}$\n\n(a) explain why $f$ has a removable…

consider the following function.\n\n$f(x)=\\frac{x - 4}{x^{2}-16}$\n\n(a) explain why $f$ has a removable discontinuity at $x = 4$. (select all that apply.)\n\n$\\square\\lim_{x\\to4}f(x)$ is finite.\n\n$\\square f(4)$ is undefined.\n\n$\\square f(4)$ and $\\lim_{x\\to4}f(x)$ are finite, but are not equal.\n\n$\\square\\lim_{x\\to4}f(x)$ does not exists.\n\n$\\square$ none of the above\n\n(b) redefine $f(4)$ so that $f$ is continuous at $x = 4$ (and thus the discontinuity is \removed\).\n\n$f(4)=\\square$\n\nresources

consider the following function.\n\n$f(x)=\\frac{x - 4}{x^{2}-16}$\n\n(a) explain why $f$ has a removable discontinuity at $x = 4$. (select all that apply.)\n\n$\\square\\lim_{x\\to4}f(x)$ is finite.\n\n$\\square f(4)$ is undefined.\n\n$\\square f(4)$ and $\\lim_{x\\to4}f(x)$ are finite, but are not equal.\n\n$\\square\\lim_{x\\to4}f(x)$ does not exists.\n\n$\\square$ none of the above\n\n(b) redefine $f(4)$ so that $f$ is continuous at $x = 4$ (and thus the discontinuity is \removed\).\n\n$f(4)=\\square$\n\nresources

Answer

Explanation:

Step1: Simplify the function

First, factor the denominator (x^{2}-16=(x + 4)(x - 4)). Then the function (f(x)=\frac{x - 4}{x^{2}-16}=\frac{x - 4}{(x + 4)(x - 4)}=\frac{1}{x + 4}), for (x\neq4).

Step2: Calculate the limit as (x\rightarrow4)

Using the simplified form (f(x)=\frac{1}{x + 4}) (for (x\neq4)), we find (\lim_{x\rightarrow4}f(x)=\lim_{x\rightarrow4}\frac{1}{x + 4}). Substitute (x = 4) into (\frac{1}{x+4}), we get (\lim_{x\rightarrow4}f(x)=\frac{1}{4 + 4}=\frac{1}{8}).

Step3: Check the value of (f(4))

The original function (f(x)=\frac{x - 4}{x^{2}-16}), when (x = 4), the denominator (x^{2}-16=4^{2}-16=0). So (f(4)) is undefined.

Answer:

(a)

  • (\lim_{x\rightarrow4}f(x)) is finite. Since (\lim_{x\rightarrow4}f(x)=\frac{1}{8})
  • (f(4)) is undefined. Because substituting (x = 4) into (f(x)=\frac{x - 4}{x^{2}-16}) makes the denominator (0)

(b) To make (f(x)) continuous at (x = 4), we re - define (f(4)) as (\lim_{x\rightarrow4}f(x)). So (f(4)=\frac{1}{8})