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I have $$f(x) = x \left| x - 1 \right|$$

Here my given value for $x$ is 1

And I need to test if the function is "continuous" when $x$ is $1$ and also when reaching $$ f(1)$$

$$ \lim\limits_{x \to 1} f(x)$$

So I plug in 1 to the equation to find out the value of $f(1)$

$$ f(1) = 1|1-1| = 0 $$ Then I check the $lim$ to see if I get $0$ as well.

$$ \lim\limits_{x \to 1} f(x) = x|x-1| = 1|1-1| = 0$$

What I am not understanding is why I need to even check for $lim$ even from the first place because the process looks identical. But I am taught to do it anyway.

Could an equation not have the same value for $f(AnyValue)$ and $\lim\limits_{x \to AnyValue} f(x)$ ?

Siddharth Bhat
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Blue Bug
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    Consider what happens with $f(x)=\begin{cases}0& x\neq0\1&x=0\end{cases}$ at $x=0$. What is $f(0)$? What is $\lim_{x\to0}f(x)$? – symplectomorphic Apr 29 '16 at 14:37
  • (You are right "the process looks identical," but that is only because your function is easy: in fact, your function is continuous, and that is why taking the limit looks the same as simply plugging in the value $x=1$. Rigorously justifying your limit requires more than simply plugging in $x=1$, and in general, taking a limit as $x\to a$ is not equivalent to simply evaluating the function at $x=a$.) – symplectomorphic Apr 29 '16 at 14:40

2 Answers2

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I think you are confused between the definition of any old function and a continuous function.

As you said, it is only for continuous functions that the fact

$$\lim_{x \to a} f(x) = f(a)$$

Any discontinuous function will not exhibit this behaviour.

As an example, consider $f(x) = \frac{1}{x}$ as $\lim{x \to 0+}$ (that is, approaching from the right hand side) versus $\lim{x \to 0-}$. I have put up the function here:

image of y = 1/x

If you look at the graph of $\frac{1}{x}$, the function is discontinuous at $x = 0$. Morever, if we approach it from the left ($\lim{x \to 0-}$), then we get the values as "$-\infty$" (this is only conventional shorthand to say that diverges off to negative infinity. It does not "reach" infinity).

However, if we look at the graph from the right hand as $\lim x \to 0+$, then we get "$+ \infty$" (which, vice versa, means that the function diverges off to positive infinity)

Now, the graph of your function looks like this: WolframAlpha plot

So, in your question, you are asked to look at $$\lim_{x \to 1} f(x) = x |x - 1|$$

As you have verified, the limit $$\lim_{x \to 1} f(x) = 1 * |1 - 1| = 0$$ is equal to the value at $$f(1) = 0$$

This means that the function is continuous.

To get a better understanding of why this is a criteria for continuity, think of what is happening. The condition is basically saying that in the smallest of regions, the value at the point $a$ of the function $f$ should be same in the smallest region around $a$. This does not give the function a lot of wiggle room, and is hence forced to be "close to each other" at small intervals, thereby ensuring what we see as continuity.

Siddharth Bhat
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The specific value matters for the function might be continuous at some values and not at others. For example $|x|/x=f(x)$ is continuous everywhere except $x=0$ for if you approach zero from the left you will get -1 but if you approach it from the right you will get 1. Also you can not just plug in x=0 for f(x) since the expression would have no meaning thus you have to check the limit.

A.Riesen
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