Evaluating limits in fractions

Question

When you want to find the limit of a fraction e.g. $\frac{1-x}{1-x^3}$ as $x$ tends to $1$. Why can you not just plug in x into the numerator and denominator?

Why do you have to make all the $x$ terms go in the denominator?

Answer 1

$$\frac{1-x}{1-x^3}=\frac{(1-x)}{(1-x)(1+x+x^2)}$$

Answer 2

Hint: Use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 3

If you plug $x=1$ into $\frac{1-x}{1-x^3}$ you get $\frac{1-1}{1-1^3}=\frac{0}{0}$ which is an indeterminate form. How do you figure that this "tends to 1"?

However, since $1-x^3=(1-x)(1+x+x^2)$, we have $$\frac{1-x}{1-x^3} = \frac{1}{1+x+x^2} \text{ for all } x \neq 1.$$ So these functions will have the same limit as $x$ approaches $1$.

Answer 4

In this particular case, you can't plug in $1$ because the function $\frac{1-x}{1-x^3}$ has a discontinuity problem at $x=1$. More specifically, you will end up with division by zero and not be able to proceed any further.

The idea of plugging the value that $x$ is approaching into the function to figure out the limit is based on the concept of continuity. The limit of a function that's continuous at a particular point is going to equal that function's value at that point:

$$\lim_{x\to x_0}f(x)=f(x_0).$$

Visually, "continuous" just means that there are no holes in the graph of a function which is not the case for $\frac{1-x}{1-x^3}$ which has got a hole at $x=1$. So, direct substitution is not going to work there. You will have to look for other ways to evaluate that limit such as finding a function that's equivalent to the original function and yet continuous at $x=1$ (the denominator is a difference of two cubes, so things can be canceled). Then, plugging in $1$ will work.