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Let $$ f(x) = \left\{ \begin{array}{ll} x^2 & \quad 0 \leq x < 1 \\ 2 & \quad 1 \leq x \end{array} \right. $$ -$f(x)$ is discontinuous at $x=1$ as it is a jump discontinuity.

-Let $$a_n = 1 + \frac{1}{n}$$

$$b_n = 1 - \frac{1}{n}$$

-For both of these sequences, as $$n \rightarrow \infty, a_n,b_n \rightarrow 1$$-It then follows that:

$$ f(a_n) = \left\{ \begin{array}{ll} (1 + \frac{1}{n})^2 & \quad 0 \leq n < 1 \\ 2 & \quad 1 \leq n \end{array} \right. $$ $$ f(b_n) = \left\{ \begin{array}{ll} (1 - \frac{1}{n})^2 & \quad 0 \leq n < 1 \\ 2 & \quad 1 \leq n \end{array} \right. $$ -Which implies that as $n\rightarrow 1,$ $f(a_n)\rightarrow 4$ and $f(b_n)\rightarrow 0$

-Since the sequences approach different points, the limit does not exist, thus, I can conclude that the function is discontinuous at $x=1$.

-Is this correct? Is this enough proof that $f(x)$ is discontinuous at $x=1$?

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5 Answers 5

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Calculate the one-sided limits at $x=1$:

  1. $\displaystyle\lim_{x\to 1+}f(x)=\lim_{x\to 1+}2=2$
  2. $\displaystyle\lim_{x\to 1-}f(x)=\lim_{x\to 1-}x^2=1$

Since they are not equal, $f$ is not continuous at $x=1$. The rest that you did has some mistakes (you do not calculate one-sided limits when you should) but more importantly is not necessary (and may be time consuming in an exam with limited time).

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  • $\begingroup$ Thank you for mentioning one-sided limits. To me, that's the easiest way to prove the discontinuity. $\endgroup$
    – shoover
    Mar 7, 2016 at 18:57
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I think you have a right idea (however possibly not the most efficient idea), but there are two "errors" in your execution of the proof.

1) Your selection of the sequences $a_n$ and $b_n$ are chosen well. One resides on the left of $x=1$, where $f(x)$ resembles $x^2$. The other resides on the right of $x=1$, where $f(x)$ resembles 2. By choosing your sequences this way, you can show that you get different behaviors when approaching $x=1$.

However, when you go and apply these sequences, you're still considering both sides of $f(x)$, thereby erasing your clever work in picking $a_n$ and $b_n$. Instead of what you have, you should simply note that

$$ f(a_n) = 2 $$ $$ f(b_n) = \left(1 - \frac{1}{n} \right)^2 $$

since $a_n > 1$ and $b_n < 1$ for all integers $n$.

2) $f(a_n) \to 2$, not 4. Also, $f(b_n) \to 1$, not 0.

EDIT: sorry, I had $a_n$ and $b_n$ flipped. Should be fixed now.

EDIT2: Just kidding, I had it right the first time, haha.

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Your original reasoning started well, but it became unclear (and errors appeared) halfway through. I think your approach was a little too long and complicated for the question that was asked. That sort of reasoning may be required when working in Metric Spaces more complicated that $\mathbb{R}$, such as $\mathbb{R}^n$, or even totally abstract Metric Spaces. It is a good reflex to use sequences, but it is a little tedious for the question here. As it has already been said, this will suffise: $$ \lim_{x\to 1^+} f(x)=2 \neq \lim_{x\to 1^-} f(x)=1.$$ Since the single-sided limits of the quantity $f(x)$ are different, the two-sided limit is ill-defined.

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The problem in your solution is that you're letting $n\to 1$ and the way you wrote $f(a_n)$ and $f(b_n)$ are not exactly right. Instead you should have $f(a_n)=2$ and $f(b_n)=(1-\frac{1}{n})^{2}$ for all $n\geq 1$. Now as $n\to\infty$ you get the desired result.

Also to your second question, note that proving discontinuity at $x=1$ is enough, and in fact that's as far as we can get as $f$ is composed of two continuous pieces that fail to merge at the point $x=1$.

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Things look OK up to and including the part where you observe that $ a_n,b_n \rightarrow 1$ as $n \rightarrow \infty$. After that, you seem to lose your train of thought.

You define cases for $f(a_n)$ and $f(b_n)$ for $0\leq n<1$, but the only integer satisfying those conditions is $0$, and at $n=0$ both $1+1/n$ and $1-1/n$ are undefined. Then, after setting things up so nicely for taking one-sided limits as $n \rightarrow \infty$, you try to take a limit as $n \rightarrow 1$, although $n$ is (presumably) still an integer.

To keep yourself on track it might pay to write (on some scrap paper, not in the proof!) a few values of $f(a_n)$ and $f(b_n)$ to see the pattern that develops: \begin{align} f(a_1) &= f\left(1+\frac11\right) = 2 & f(b_1) &= f\left(1-\frac11\right) = 0 \\ f(a_2) &= f\left(1+\frac12\right) = 2 & f(b_2) &= f\left(1-\frac12\right) = \left(1-\frac12\right)^2 \\ f(a_3) &= f\left(1+\frac13\right) = 2 & f(b_3) &= f\left(1-\frac13\right) = \left(1-\frac13\right)^2 \\ f(a_4) &= f\left(1+\frac14\right) = 2 & f(b_4) &= f\left(1-\frac14\right) = \left(1-\frac14\right)^2 \\ &\quad\vdots &&\quad\vdots \end{align}

I avoided fully evaluating the $f(b_n)$ values to $1/4$, $4/9$, $9/16$, etc. because it seems a little easier to see that they are converging to $1$ if you leave them in the form $(1-1/n)^2$.

With a few examples like this, it might be easier to remember that $n$ is just the subscript of the thing you're applying $f$ to, not something to directly apply $f$ to, and it might also be easier to see that the left- and right-side limits are $1$ and $2$, respectively.

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