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If I have a hole in a rational function, but the hole is located at a vertical asymptote, is the hole still recognized? For example in the equation (x+3)(x+1)/(x+1)(x+1) I will have a hole at -1, but a vertical asymptote also at -1.

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1 Answer

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A "hole" is really just a point where the function is undefined, but it just so happens that the function "looks smooth" around the point. This is called a removable discontinuity, and is defined as a point $c$ so that $\lim_{x\to c}f(x)$ exists and is finite, but $f(c)$ is undefined. By contrast, an asymptote occurs when the left and right hand limits are individually $\pm\infty$ and in particular not finite. Now let's take your example: $f(x)=(x+1)/(x+1)^2$, and take the limit as $x$ goes to $-1$. There's no hole here, since the limits go to $\pm\infty$ and is not finite. So we don't draw a "hole", but just the asymptote instead!

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