Calculus problems

Solution (short): We have

One can check that

if

then $\dfrac{x-2}{x^2-4}\approx .2499937502$
if

then $\dfrac{x-2}{x^2-4}\approx .2500062502$

Both are very close to

(which is

). This suggests that

$\lim\limits_{x \to 2} \dfrac{x-2}{x^2-4}=\dfrac{1}{4}$

Compare

with

Note that in both cases we have

indeterminate forms. The second limit is 1, since

In other words, the zero term

gets canceled and the limit perfectly exists as a finite number. To apply the same idea to

we first factor the bottom and then cancel the zero term

We first note that we can not use the Direct Substitution Property to evaluate $\lim\limits_{x \to 2} \dfrac{x-2}{x^2-4}$ , since the denominator

is 0 at

. In other words, 2 is not from the dom ain of $\dfrac{x-2}{x^2-4}$ , therefore DSP can not be used. However, this does not necessarily mean that the limit does not exist. Actually, this limit does exist.

To understand it, we first note that the problem of finding the limit $\lim\limits_{x \to 2} \dfrac{x-2}{x^2-4}$ is different from the problem of evaluating $\dfrac{x-2}{x^2-4}$ at

The latter is impossible, since we can't divide by 0. Nevertheless, the limit of $\dfrac{x-2}{x^2-4}$ when

is goes to 2 does exist. Indeed, by the nature of a limit when we are evaluating $\lim\limits_{x \to 2} \dfrac{x-2}{x^2-4}$ we are basically trying to see what happens to $\dfrac{x-2}{x^2-4}$ when

is close to 2 (as close to 2 as we like), but not equal to 2 (see Definition of limit). One can check that

then $\dfrac{x-2}{x^2-4}\approx .2499937502$
if

then $\dfrac{x-2}{x^2-4}\approx .2500062502$

Both are very close to

(which is

). This suggests that

$\lim\limits_{x \to 2} \dfrac{x-2}{x^2-4}=\dfrac{1}{4}$

However, our goal here is to prove it rigorously using the Limit Laws. To this end, we note that the numerator of $\dfrac{x-2}{x^2-4}$ is also 0 when

In other words, when

approaches 2, the top and the bottom of $\dfrac{x-2}{x^2-4}$ go to zero. In this case we say that the limit is a $\dfrac{0}{0}$ indeterminate form. To evaluate this lim it, we note that since $x\not = 2$ (as we discussed before), then, by algebra, we have

$\dfrac{x-2}{x^2-4} \stackrel{A1}{=}\dfrac{x-2}{x^2-2^2} \stackrel{A2}{=}\dfrac{x-2}{(x-2)(x+2)} \stackrel{A3}{=}\dfrac{(x-2)\cdot 1}{(x-2)(x+2)} \stackrel{A4}{=}\dfrac{1}{x+2}$

STEP A1: We show that $\dfrac{x-2}{x^2-4}=\dfrac{x-2}{x^2-2^2}$ .

WORK: By arith me tic

and $2\cdot 2=2^2.$ (by definition of second power). Therefore

$4=2\cdot 2=2^2$

hence

$\dfrac{x-2}{x^2-4}=\dfrac{x-2}{x^2-2^2}$

STEP A2: We show that $\dfrac{x-2}{x^2-2^2}=\dfrac{x-2}{(x-2)(x+2)}$ .

WORK: Here we use the identity

In our case

and

therefore

hence

$\dfrac{x-2}{x^2-2^2}=\dfrac{x-2}{(x-2)(x+2)}$

STEP A3: We show that $\dfrac{x-2}{(x-2)(x+2)}=\dfrac{(x-2)\cdot 1}{(x-2)(x+2)}$ .

WORK: Multiplication by 1 does not change the expression. Therefore