1
$\begingroup$

we have been doing epsilon and delta limit proofs in my first year calculus class and I can't seem to wrap my around this stuff. We have not done an example in class of disproving a limit, and this was given as a practice question for my exam so I really need some help, Thank You!

Prove that the $\lim_{x \rightarrow 0}\frac1x$ does not exist

$\endgroup$
2
  • 2
    $\begingroup$ $$\lim_{x\to0^+}\frac1x=+\infty, \lim_{x\to0^-}\frac1x=-\infty$$ $\endgroup$
    – lab bhattacharjee
    Dec 10, 2013 at 10:56
  • 1
    $\begingroup$ Yeah I unnderstand that, but I need to prove it using the definition of a limit (epsilons and deltas) Thanks! $\endgroup$
    – Alex Chavez
    Dec 10, 2013 at 10:57

1 Answer 1

Reset to default
4
$\begingroup$

Show that no number $\alpha \in \mathbb R$ is a limit of $\frac 1x$ for $x \to 0$. That is show that $$ \exists \epsilon > 0 \; \forall \delta > 0 \; \exists x:\; x \in (-\delta, \delta) \land\left|\frac 1x - \alpha\right| > \epsilon $$ Let $\epsilon = 1$ and given $\delta$, set $x = \min\{\frac\delta 2, \frac 1{|\alpha| + 2}\}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .