# Thread: How to prove that a limit exists at 0.

1. ## How to prove that a limit exists at 0.

Which of the following functions with domain R-{0} have a limit at 0 and what is it?

(1) f(x)=x

(2) f(x)=1/x

How would I prove these either using the definition of limit, the Sequential Criterion for limits, or the Squeeze thm for limits of functions?

For number (1), is this on the right track?

Let $\epsilon$>0 be given. Suppose that $\delta$= $\epsilon$. Then if |x|< $\epsilon$, |f(x)|=|x|< $\epsilon$ is true.

Is that correct for number 1? and how would I go about proving #2? I know that it goes to 0 because of the squeeze theorem but how do I write a formal proof for that?
Thanks

2. Originally Posted by alice8675309
Which of the following functions with domain R-{0} have a limit at 0 and what is it?

(1) f(x)=x

(2) f(x)=1/x

How would I prove these either using the definition of limit, the Sequential Criterion for limits, or the Squeeze thm for limits of functions?

For number (1), is this on the right track?

Let $\epsilon$>0 be given. Suppose that $\delta$= $\epsilon$. Then if |x|< $\epsilon$, |f(x)|=|x|< $\epsilon$ is true.

Is that correct for number 1? and how would I go about proving #2? I know that it goes to 0 because of the squeeze theorem but how do I write a formal proof for that?
Thanks
You are correct for the first one. However the limit of the 2nd one does not exist.

The limit from the right is $\infty$ and the limit from the left is $-\infty$

for example let $x_n \to 0$ as $n \to \infty$ and further suppose that $x_n > 0$ for each n. Now Let $M \in \mathbb{R}^+$This gives

$\displaystyle f(x_n)=\frac{1}{x_n}$ Now choose $N \in \mathbb{Z^+}$ such that for $n > N$ $\displaystyle x_n < \frac{1}{M}$ this would imply that $\displaystyle f(x_n) > M$ since M is arbitrary the sequence is unbounded.

3. Originally Posted by TheEmptySet
You are correct for the first one. However the limit of the 2nd one does not exist.

The limit from the right is $\infty$ and the limit from the left is $-\infty$

for example let $x_n \to 0$ as $n \to \infty$ and further suppose that $x_n > 0$ for each n. Now Let $M \in \mathbb{R}^+$This gives

$\displaystyle f(x_n)=\frac{1}{x_n}$ Now choose $N \in \mathbb{Z^+}$ such that for $n > N$ $\displaystyle x_n < \frac{1}{M}$ this would imply that $\displaystyle f(x_n) > M$ since M is arbitrary the sequence is unbounded.

Thanks so much. However, i think i meant to put this up as my number two: f(x)=xsin(1/x) which by then is 0 by the squeeze thm for limits of functions. However, how would i put this into a proof?

4. Originally Posted by alice8675309
Thanks so much. However, i think i meant to put this up as my number two: f(x)=xsin(1/x) which by then is 0 by the squeeze thm for limits of functions. However, how would i put this into a proof?
Notice that

$\displaystyle -1 \le \sin\left( \frac{1}{x}\right) \le 1 \implies -x \le \bigg|\sin\left( \frac{1}{x}\right) \bigg| \le x \implies$

Or $\displaystyle \bigg|\sin\left( \frac{1}{x}\right) \bigg| < |x|$

so as you said use the squeeze theorem and the same delta and epsilon will work that you used in the first part