How to find this limit: The question is to find this limit without using H.R (Hopital Rule):
 $$\lim_{x\to 1}\frac{x^4 -1}{x^3 -1}$$
So this will be $\frac{0}{0}$
Which is indetermined but using H.R we find it is 
 $\frac{4}{3}$ 
But I need another method to get the answer so how could I get it?
 A: You should decompose the two polynomial and the result will be very easy:
$$
\lim_{x\rightarrow 1} \frac{x^4-1}{x^3-1}=\lim_{x\rightarrow 1} \frac{(x-1)(x+1)(x^2+1)}{(x-1)(x^2+x+1)}=\lim_{x\rightarrow 1} \frac{(x+1)(x^2+1)}{(x^2+x+1)}=\frac{4}{3}
$$
A: Since good answers have been given, I will permit myself the luxury of a less good one.
It is easier (or at least more familiar) to see what happens as a variable approaches $0$. So let $x=1+t$.  Imagine expanding $(1+t)^4$ using the Binomial Theorem. We get $1+4t+ \text{terms in higher powers of $t$}$. Subtract $1$. We get
$$4t+ \text{terms in higher powers of $t$}.$$
Do something similar with the bottom. We get 
$$3t+ \text{terms in higher powers of $t$}.$$
Divide, and cancel a $t$. We get
$$\frac{4+ \text{terms in $t$ and higher powers of $t$}}{3+ \text{terms in $t$ and higher powers of $t$}}.$$
As $t\to 0$, the parts that involve positive powers of $t$ die, and our limit is $\frac{4}{3}$.
A: just to expound on Ikki's answer, any polynomial of the form $x^n-1$ can be written as $$(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$
The second factor is called a cyclotomic polynomial.  Thus you can remove the discontinuity in the denominator since $(x-1)$'s will cancel.
