Compute the following limits (think about the {theorems} you
use).
\begin{alignat*}{3}
&a) \lim_{x\to 1} \frac{x^3-1 }{ (x-1)^2} & &b) \lim_{x\to{-2}}
\frac{x^3 + 8 }{ x+2} & &c) \lim_{x \to 1} \frac{x^3-2x^2 + 2x -1 }{
x^3 -1} \\
&d) \lim_{x\to 8} \frac{\root 3 \of x - 2 }{ x-8} & &e) \lim_{x\to{-1}}
\frac{\frac{1}{ x}+ 1}{x+1} & &f) \lim_{x \to a} \frac{x^n - a^n }{ x-a} \\
&g) \lim_{x\to 2} \frac{\sqrt 2 - \sqrt x }{ 2-x} & &h) \lim_{x \to 0}\frac{1-
\sqrt{1-x^2} }{ x^2} & &i) \lim_{x \to 4} \frac{\root 3 \of x - \root 3
\of 4 }{ x-4}
\end{alignat*}
The “theorems” you are supposed to think about are the limit laws that justify plugging in after the fraction has been simplified. If I were giving the problem, I’d probably be inclined not to call them theorems, but rather give an intuitive explanation for why a removable discontinuity can be filled in with the value obtained by plugging into the simplified fraction. On the other hand, most students haven’t complained about this point because (a) it’s intuitive, (b) it’s easy. The combination means that only truly rigorous students (the ones who ask “but WHY is that true?”) will notice something needs justifying.
– Eric Hsu 1998-08-09