toundation for success in calculus.

Consider the function $f$ defined by $f(x)=\frac{x^4-1}{x-1}.$

a$.$ By successive evaluation of $f$ at $x=1\mathrm{.}8,1\mathrm{.}9,1\mathrm{.}99,1\mathrm{.}999,$ and $1\mathrm{.}9999,$ what do you think happens to the values of $f$ as $x$ increases towards $2?$

b$.$ Do a similar experiment on for values of $x$ slightly greater than $2.$ Again, comment on your results.

As a shorthand, and anticipating a forthcoring definition, we shall describe what you found in parts a and $b$ by witing $\underset{x2}{lim}f(x)=15,$ or more specifically, $\underset{x2}{lim}\frac{x^4-1}{x-1}=15$

c$.$ In this particular case you could have "cheated" by immediately evaluating $f$ at $2.$ Get a computer plot of the function between $1\mathrm{.}8$ and $2\mathrm{.}2$ to illustrate what happens in this straightforward situation.

$2.$ Use the same function $f$ as above, but this time consider what happens as $x$ approaches $1.$

a$.$ Study this situation experimentally as you did in parts a and $3$ of Problem $1.$ To gain some additional feel and respect for the situation, compute the numerator and denominator of $f$ separately for several $x$ values before dividing. What are your conclusions and, in particular, what is $\underset{x1}{lim}f(x)?$

b$.$ What happens when you try to "cheat" as was done in part $c$ of Problem $1?$ There are situations in which direct evaluation at the specified point is possible and actually gives the limit$.$ These give rise to a concept called continuity. There are many important situations in calculus when this technique will not work, however.