# Evaluating limits in fractions

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?

• What do you mean by "make all the $x$ terms go in the denominator" – Peter Foreman Apr 21 at 19:05

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 $$1$$.

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

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.

• Thank you all. So if you plug in the value which leads the denominator to be 0 (i.e. making the function undefined), then you must rearrange. – PhysicsEnthu Apr 21 at 19:52
• Right. Try to manipulate the function algebraically to find an equivalent function that has no continuity problem at that point and use the property $\lim_{x\to x_0}f(x)=f(x_0)$. – Michael Rybkin Apr 21 at 20:27

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