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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?

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  • $\begingroup$ What do you mean by "make all the $x$ terms go in the denominator" $\endgroup$ – Peter Foreman Apr 21 at 19:05
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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 $1$.

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

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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.

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  • $\begingroup$ 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. $\endgroup$ – PhysicsEnthu Apr 21 at 19:52
  • $\begingroup$ Right. Try to manipulate the function algebraically to find an equivalent function that has no continuity problem at that point and use the property $\lim_{x\to x_0}f(x)=f(x_0)$. $\endgroup$ – Michael Rybkin Apr 21 at 20:27
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Hint: Use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

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