University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises - Page 109: 73


$$f(x)=\frac{2}{(x-2)^2}$$ The graph is below.

Work Step by Step

(What I present here is just an example. There are other possible functions which satisfy the exercise.) - As $\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=\infty$, I would suppose that my function would have this rational form $A/B$, and in the denominator $B=(x-2)^2$ to make the function approach $\infty$ as $x\to2^-$ and $2^+$. Also $A\gt0$ to have the necessary result $\infty$. $$f(x)=\frac{A}{(x-2)^2}$$ - As $\lim_{x\to\pm\infty}f(x)=0$: $$\lim_{x\pm\infty}\frac{A}{(x-2)^2}=\lim_{x\pm\infty}\frac{A}{x^2-2x+1}=0$$ Remember the method that divides both numerator and denominator by the highest degree of $x$ in the denominator, which is $x^2$ in this case: $$\lim_{x\pm\infty}\frac{\frac{A}{x^2}}{1-\frac{2}{x}+\frac{1}{x^2}}=0$$ $$\frac{\lim_{x\pm\infty}\frac{A}{x^2}}{1-0+0}=\lim_{x\pm\infty}\frac{A}{x^2}=0$$ We know that as $A\in Z$, $\lim_{x\to\pm\infty}(A/x^2)=0$, so we can choose any positive number $(A\gt0)$ here. I would choose $A=2$. So in conclusion, the function I come up with is $$f(x)=\frac{2}{(x-2)^2}$$. The graph is shown below.
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