The equation $$\frac{(x^2 + 4x - 32)}{(x^2 - 8x + 16)}$$ factors to $$\frac{(x + 8)(x - 4)}{(x - 4)(x - 4)}$$ This can be simplified to $$\frac{(x + 8)}{(x - 4)}, x \neq 4$$

I am being told, on Khan Academy, that there is a vertical asymptote at $x = 4$. I'm not sure how this can be. The simplified expression only holds in the domain $x \neq 4$, and the non-simplified expression evaluates to $\frac{0}0$, when $x = 4$. So far as I'm aware, $\frac{0}0$ is indeterminate, and as such should indicate there is a discontinuity at that value of $x$.

Most likely Khan Academy is right, and there's some hole in my knowledge. Could someone help me out with why this function has a vertical asymptote at $x = 4$ rather than a discontinuity?

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    $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Sep 21 '17 at 15:23
  • $\begingroup$ Because your simplification is wrong: It is $(x + 8) / (x - 4)$, which does have an asymptote (if $x \approx 4$, the denominator is very close to zero). $\endgroup$ – user296602 Sep 21 '17 at 15:25

There is a vertical asymptote at $x=4$. Observe that $$\lim_{x \to 4^+} \frac{x+8}{x-4} = \infty, \lim_{x \to 4^-}\frac{x+8}{x-4}=-\infty,$$ meaning there is an asymptote at $x=4$. To informally justify this to yourself, try plugging in values very close to 4; if the denominator gets small and the numerator isn't changing much, the fraction will blow up or down to $+\infty$ or $-\infty$. If it were a removable discontinuity, $\lim_{x\to4}f(x)$ would exist.

  • $\begingroup$ So if I understand the logic, it's that because there is an asymptote at 4, x will never equal 4, and therefore there can be no discontinuity? Something like that? $\endgroup$ – NotAnAmbiTurner Sep 22 '17 at 4:55
  • $\begingroup$ The function is not defined at 4, so you're not going to be able to 'plug in' 4 into the function. The function is still definitely discontinuous at $x=4$. The limiting behavior, in other words, what the function does for $x$ values near 4, dictates the shape of the graph around that discontinuity. For a vertical asymptote, the graph will be shooting upwards or downwards towards $\infty$ or $-\infty$. $\endgroup$ – m_squared Sep 22 '17 at 13:13
  • $\begingroup$ The domain of the function $$f(x) = \frac{x^2 + 4x - 32}{x^2 - 8x + 16} = \frac{(x + 8)(x - 4)}{(x - 4)(x - 4)}$$ is all real numbers except $4$. In this case, canceling a factor of $x - 4$ from the numerator and denominator does not change the domain. The function is continuous everywhere on its domain. It has a vertical asymptote at $x = 4$ for the reasons stated in m_squared's answer. $\endgroup$ – N. F. Taussig Sep 23 '17 at 10:23
  • $\begingroup$ @m_squared that's really helpful. Didn't occur to me to think of an asymptote as type of discontinuity, but it obviously is. Thanks! :) $\endgroup$ – NotAnAmbiTurner Sep 30 '17 at 22:31

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