How to prove that a limit exists at 0.

**alice8675309** · Mar 27th 2011, 11:28 AM

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

**TheEmptySet** · Mar 27th 2011, 12:56 PM

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.

**alice8675309** · Mar 27th 2011, 02:13 PM

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?

**TheEmptySet** · Mar 27th 2011, 02:22 PM

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

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