Indeterminate and 2 as answer to $\frac {x^2-1}{x-1}$ at x = 1

Question

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

When we directly put x=1 to $\frac {x^2-1}{x-1}$ without simplifying we get $\frac{0}{0}$ indeterminate form but on simplifying further $\frac {x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1}=x+1$ and then we put x=1, we get value as 2.

So is it right to say $\frac {x^2-1}{x-1}$ is indeterminate and 2 as answer at x=1?

Answer 1

It would be right to say that it is an indeterminant form $(0/0)$ as $x=1$ is not in the domain of the function and the limiting value of the function is 2 as $x$ tends to 1 (not when equal) .

Answer 2

The function $f(x)$ is not defined at $x = 1$. However, there exist a continuous extension $g(x) = x+1$, defined in all $\mathbb{R}$.

Answer 3

There is an underlying issue. Whether you can avoid regarding the fraction as indeterminate in the first place. Yes, because the limit of a product is the product of limits. So, the problem can be re-expressed as

$$\left[\lim_{x \to 1} \frac{x-1}{x-1}\right] \times \left[\lim_{x\to 1} (x + 1) \right]. \tag{A} $$

Although L'Hopital's Rule can (if you wish) be applied to the first factor in (A) above, it is not really necessary. That is, as $x$ approaches, but never equals $(1)$, it is clear that the first factor represents the constant value of $(1)$.

After a while, you get a feel of whether you can shortcut the process by simply cancelling terms. Generally, yes, because (when two limits are each convergent), the product of the limits equals the limit of the product.