Numerically evaluating the limit of $\frac{x^4-1}{x^3-1}$ as $x\rightarrow 1$ What is the limit as $x \to 1$ of the function 
$$ f(x) = \frac{x^4-1}{x^3-1} . $$
 A: Generally, the "numerical method" for finding limits involves substituting numbers closer and closer to the number and seeing if we get a pattern. 
In this example, we have $f(x)=\dfrac{x^4-1}{x^3-1}$. We should determine $f(1.1), f(1.01), f(0.99), f(1.001), f(0.999), \dots$ and see if a pattern develops. If these numbers approach some number, that would be our numerical estimate of the limit.
A: If we just divide, we get $f(x) = x+ \frac{1}{1+x+x^2}$, from which the limit easily follows, numerically or otherwise.
A: Applying L'Hôpital's rule makes it quite straightforward.
$$ \lim_{x \rightarrow 1}\; \frac{x^4-1}{x^3-1} = \lim_{x \rightarrow 1} \;\frac{4x^3}{3x^2} =  \lim_{x \rightarrow 1} \;\frac{4}{3}x = \frac{4}{3}$$
A: Note that $x^4-1=(x-1)(x^3+x^2+x+1)$ and $x^3-1=(x-1)(x^2+x+1)$.
Thus
$$f(x)=\frac{(x-1)(x^3+x^2+x+1)}{(x-1)(x^2+x+1)}.$$
When $x\ne 1$, the $x-1$ terms cancel. Now we can safely let $x$ approach $1$.
A: A crucial fact from algebra should be remembered: If you plug $x=1$ into a polynomial and get $0$, that tells you that $(x-1)$ is one of the factors.  (Likewise if you plug in $x=9$ and get $0$, then $(x-9)$ is one of the factors, etc.)  Thus
$$
\frac{x^4-1}{x^3-1} = \frac{(x-1)(\cdots\cdots\cdots)}{(x-1)(\cdots\cdots\cdots)}.
$$
So cancel $(x-1)$ from the top and the bottom and go on from there.  (To find the expressions to put in place of "$(\cdots\cdots\cdots)$", you can use long division if all else fails.)
A: The most straightforward approach probably involves the fact that $x^{n}-1 = (x-1)(x^{n-1}+\dots+x+1)$, which you can verify by multiplying a few of the terms and noticing the pattern. If you rewrite the numerator and denominator using this relation, there will be cancellation of the term $(x-1)$, which is okay since $x-1=0$ only when $x$ is $1$, itself, an we are taking a limit as $x\rightarrow 1$.
Alternatively, If you have learned L'Hopital's Rule, then since "plugging in 1 for x" yields an indeterminate form ($\frac{0}{0}$), 
$$\lim_{x\rightarrow 1}\frac{x^{4}-1}{x^{3}-1} = \lim_{x\rightarrow 1}\frac{\frac{d}{dx}x^{4}-1}{\frac{d}{dx}x^{3}-1}$$
$$= \lim_{x\rightarrow 1}\frac{4x^{3}}{3x^{2}} = \lim_{x\rightarrow 1}\frac{4x}{3} = \frac{4}{3}$$
