Simplify $\dfrac{x^2-x}{x^2-1}$

Question

My task is to calculate the following: $$\lim_{x\to1}\dfrac{x^2-x}{x^2-1}$$

My attempt was to simplify

$$\dfrac{x^2-x}{x^2-1}$$

But this is where I am stuck. I tried polynomial division but I did not get a nice looking result.

However wolfram alpha showed me this solution (for $x$ not equal to $1$)

$$\dfrac{x^2-x}{x^2-1}=\dfrac{x}{x+1}$$

I'd like to know why this works and how the solution can be obtained.

Answer 1

you have $x^2-x=x\color{red}{(x-1)}$ and $x^2-1 = (x+1)\color{red}{(x-1)}$

$$\dfrac{x^2-x}{x^2-1} =\dfrac{x\color{red}{(x-1)}}{(x+1)\color{red}{(x-1)}}=\dfrac{x}{x+1}$$ $$\lim_{x\to1}\dfrac{x^2-x}{x^2-x} =\lim_{x\to1}\dfrac{x}{x+1}= \dfrac{1}{2}$$

Answer 2

$$\frac{x^2-x}{x^2-1}=\frac{x(x-1)}{(x+1)(x-1)}=\frac{x}{x+1}$$

Answer 3

You simply need to factor out the $x$ on the top and expand the difference of squares on the bottom (follow this link if you don't know what a difference of squares is). Then, there will be something that you can cancel out which will help you simplify the expression. After that, you can use the direct substitution method to finally find the limit. Here are the steps:

\begin{align}\require{cancel} \lim_{x\to1}\frac{x^2-x}{x^2-1} &=\lim_{x\to1}\frac{x^2-x}{x^2-1^2}\\ &=\lim_{x\to1}\frac{x(x-1)}{(x-1)(x+1)}\\ &=\lim_{x\to1}\frac{x\cancel{(x-1)}}{\cancel{(x-1)}(x+1)}\\ &=\lim_{x\to1}\frac{x}{x+1}\\ &=\frac{1}{1+1}\\ &=\frac{1}{2} \end{align}

Answer 4

Hint:

Let $x=1+h\implies h\to0$

$$x^2-1=h(2+h)$$

$$x^2-x=(1+h)^2-(1+h)=h(1+h)$$

Now $h$ can be cancelled out safely from numerator & denominator as $h\ne0$ as $h\to0$