0
$\begingroup$

The definition of $\lim_{x\to 0} f(x) = \infty$ is:

$\forall N\in\mathbb{R}\;\exists \delta\gt 0\; (0<\lvert x\rvert<\delta\implies f(x)>N)$

The definition of "$\lim_{x\to 0} f(x)$ does not exist" is:

$\forall L\in\mathbb{R}\; \exists \epsilon\gt 0\; \forall \delta\gt 0 \; \exists x\in \mathbb{R}\; (0<\lvert x\rvert<\delta\; and\; \lvert f(x)-L \rvert\gt\epsilon)$

The question makes sense intuitively and geometrically, but how do I prove directly from the definitions that the first definition implies the second?

Thank you!

$\endgroup$
0
$\begingroup$

Suppose you had $\lim_{x\to 0} f(x) = L < \infty$ and $\lim_{x\to 0} f(x) = \infty$. Then from the first, applying it with $\epsilon := 1$, you can find $\delta > 0$ such that $|f(x) - L| < 1$ whenever $0 < |x| < \delta$. Similarly, from the second, you can find $\delta' > 0$ such that $f(x) > L + 1$ whenever $0 < |x| < \delta'$.

Now, can you see from here how to reach a contradiction?

$\endgroup$
0
$\begingroup$

Assuming $$\forall N\in\mathbb{R}\;\exists \delta\gt 0\; (0<\lvert x\rvert<\delta\implies f(x)>N)$$

You want to show that

$$\forall L\in\mathbb{R}\; \exists \epsilon\gt 0\; \forall \delta\gt 0 \; \exists x\in \mathbb{R}\; (0<\lvert x\rvert<\delta\; and\; \lvert f(x)-L \rvert\lt\epsilon)$$ is not true.

Suppose your sequence has a limit $L \in R$

Let $N=L+2\epsilon$ where the $\epsilon$ comes from the definition of limit. For this $N$ you have a $\delta $ such that $$(0<\lvert x\rvert<\delta\implies f(x)>N=L+2\epsilon)$$ That is $|f(x)-L|>2\epsilon$ which contradicts, $|f(x)-L|<\epsilon$

$\endgroup$
0
$\begingroup$

Let $f:D\rightarrow R$ be a finction and $\lim_{x\rightarrow 0} f(x)= \infty$.Now if possible let $\lim_{x\rightarrow 0}f(x)$ exists and equals to $L$, then we have a $\delta>0$ for the positive number $1$ such that whenever $0<|x|<\delta ,x\in D$ then $|f(x)-L|<1$ i.e. for $x\in (-\delta,+\delta)\cap D,x\not =0,$ we have $|f(x)|-|L|≤||f(x)|-|L||≤|f(x)-L|<1$, so that on $\{(-\delta,0)\cup (0,+\delta)\}\cap D$ we show $f$ is bounded by $|L|+1$

But by assumption $\lim_{x\rightarrow 0} f(x)= \infty$ i.e. given any $G>0$ we have $\delta>0$ such that whenever $0<|x|<\delta ,x\in D$ we have $f(x)>G$ i.e. one can choose $\{x_n\} $ from $D$ such that $x_n\rightarrow 0$ but $f(x_n)>n$ i.e. $f$ is unbounded in every deleted nbd of $0$.

Hence $\lim_{x\rightarrow 0} f(x)$ doesn't exist.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.