Calculus: $\lim_{x \to 1} \frac{x^3 - 1}{x-1}$

Question

Is the limit $0$ or $3$?

$x^3 -1$ can be $(x-1)(x^2 +x +1)$, with the transformation, the limit will be $3$? Why cannot we just say $x$ is to be $1$, $x^3 -1$ is going to be $0$ so the limit is $0$?

Answer 1

You have shown that you can see why the limit of the expression should be $3$. Regarding why it shouldn't be $0$, consider how $\displaystyle \frac{x^3-1}{x-1}=\left(x^2+x+1\right)\frac{x-1}{x-1}$ would be evaluated at different $x$-values. At $x=3$, we have $\require{cancel}\displaystyle \left(3^2+3+1\right)\frac{\cancel{2}}{\cancel{2}}$ and likewise, at $x=0.5$, we would have $\require{cancel}\displaystyle \left(0.5^2+0.5+1\right)\frac{\cancel{0.5}}{\cancel{0.5}}$. The pattern should be clear; for any $x\ne1$, the terms in the fraction are nonzero and can cancel with no problem. But this includes all $x$ in the vicinity of $1$, which is exactly what we pass through as we approach $1$. Since a limit is not interested in what happens at $x=1$ but in what happens near $x=1$, and we never need to hit $1$ exactly, we never need to evaluate $\cfrac00$ and it does not affect the answer.

Answer 2

Use $$\frac{x^3-1}{x-1}=x^2+x+1$$ for $x\neq1$.

Also, by the definition of the limit if $x$ is closed to $1$, so $x\neq1$.

Answer 3

Since x can not be 1，and x^3-1=(x-1)(x^2+x+1) ,we can transform the former one to x^2 + x + 1. The limit of it is 1+1+1=3. For why it can’t be 0. The limit of (x-1) is also 0. We can’t make a conclusion that how much is the limit of 0/0.

Answer 4

Because if = 0 then we would get 0/0 which is meaningless but the question asked is if x gets closer and closer to 1 then what does $\frac{x^3-1}{x-1}$ approach that is of hence when we say that the answer is 3 then it means that $f(x) =\frac{x^3-1}{x-1}$ can be made closer and closer to 3 by making $x$ become closer and closer to one without ever actually putting $x=1$

Answer 5

Suppose $f:\mathbb R \to \mathbb R$. Then, $\lim_{x\to 1}f(x)$ means to evaluate the limit of $f(x)$ as $x$ approaches $1$. The symbol "$\to$" inside the limit means to evaluate the value of $x$ as $x$ approaches one. This is not the same as evaluating the value of $x$ as $x$ equals one. Therefore, you can factor and cancel out the $(x-1)$ in the numerator and denominator (as $x$ approaches $1$ does not imply that $x=1$) to form

$$\lim_{x\to 1}x^2+x+1$$

Answer 6

Welcome to Math Stack Exchange! Since our initial condition for this fraction to exist is that x≠ 1, since (x-1) is in the denominator.