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Consider the following equation:

$\left(x^{2}-1\right)y=1-x$

Setting $y{=}0$ forces the LHS equal to zero, so $x$ must be 1 for the RHS to be zero too. However, if we now rearrange the equation as,

$y=\frac{1-x}{x^2 - 1}$,

and we apply L'Hopital's rule, we see that when $x{=}1$, $y{=}{-}1/2$. We see also, that y is never equal to zero. So which is it? This seems contradictory to me. Please try to be thorough with your answer for me to understand.

Anon
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  • $(1,0)$ is a solution of $\left(x^{2}-1\right)y=1-x$ and the only one with $y=0$ but not the only solution with $x=1$. For example $(1,-\frac12)$ is also a solution and in fact the only one with $y=-\frac12$. Every other value of $y$ leads to two solutions. – Henry Aug 01 '23 at 00:58

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The equations $$(x^2-1)y = 1-x \tag{1}$$ and $$y=\frac{1-x}{x^2 - 1} \tag{2}$$ are not equivalent. Each solution $(x,y)$ of $(2)$ is of course also a solution of $(1)$, but not conversely because the RHS of $(2)$ is undefined for $x = 1$. You correctly determine the limit $$\lim_{x\to 1} \frac{1-x}{x^2 - 1} = - \frac 1 2$$ and in fact $(1,- \frac 1 2 )$ is a solution of $(1)$. However, all $(1,y)$ with $y \in \mathbb R$ are solutions of $(1)$, thus there is no contradicton to the fact that $(1,0)$ solves $(1)$. I guess you erroneously believed that for each $x$ there must be a unique $y$ such that $(x,y)$ solves $(1)$.

Also note that you do not need L'Hopital's rule because $$\frac{1-x}{x^2 - 1} = - \frac{1}{x+1} $$ for $x \ne 1$.

Paul Frost
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There's no need to bring calculus into this purely algebraic question.

Dividing both sides of an equation by an expression is only valid when that expression is nonzero. So your rearrangement loses one of the solutions.

I suggest setting one side of the equation to zero and factoring: $$ (x^2-1)y = 1-x \iff (x-1)(xy + y + 1) = 0 $$ To satisfy this equation, one of the factors must be zero. If $x=1$, the equation is satisfied for any $y$ (including $y=0$). The second factor is zero if $y = - \frac{1}{x+1}$, which works as long as $x \neq -1$. The solution set is plotted below.

plot of solution set

Matthew Leingang
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L Hopital's Rule tells you the behaviour of that function around $x=1$ not exactly on $x=1$.

The latter calculates the limit, and limit does not depend on the actual value of the function at that point. It is the behaviour of the function in the neighbourhood of that point.

Suppose we define a function $f:\mathbb{R}\to\mathbb{R}$ like, $f(x)=x$ when $x\ne1$ and $f(x)=1000$ when $x=1$.

Now what is the answer of $$\lim_{x\to1}f(x)=??$$

Also, both functions that you wrote are not same.

MathStackexchangeIsNotSoBad
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