Prove using esplion-delta.

$$\lim_{x \to 1-} \left[\frac{1}{x}\right]=1$$

My prove :

All $\varepsilon>0$ exist $\delta>0$ so all x that appiles $1-\delta<x<1$ appiles $\left|\left[\frac{1}{x}\right]-1\right|<\varepsilon$

So for $\delta=1$ exist $\varepsilon>0$ .

Therefore :

$$1-\delta<x<1$$ $$1-1<x<1$$ $$0<x<1$$

As well :


Therefore :


I'm not sure I did it correctly, Since I didn't come up with $\delta$ that depends on $\varepsilon$ could somebody explain? any help will be appreciated.


First of all, if $0 < x < 1$, then you have $1 < \frac{1}{x} < \infty$, not $0 < \frac{1}{x} < 1$.

Second, you don't necessarily have to find a $\delta$ that depends on $\epsilon$. The $\epsilon$-$\delta$ definition of the limit requires that for any $\epsilon > 0$ you find a $\delta > 0$. Usually, the value of $\delta$ will change with $\epsilon$, but not always.

Notice that if $\frac{1}{2} < x < 1$, then $1 < \frac{1}{x} < 2$. Therefore, $\left[\frac{1}{x}\right] = 1$ for all $x$ such that $\tfrac{1}{2} < x < 1$.

Use that to find a $\delta$ such that $\left|\left[\frac{1}{x}\right]-1\right| < \epsilon$ for all $x$ such that $1-\delta < x < 1$.

  • $\begingroup$ If we choose $\delta=\frac{1}{2}$ therefore $$|[\frac{1}{x}]-1=|1-1|=0<\varepsilon$$, If I understood the concept right, since $\varepsilon$ doesn't rely on $\delta$ it means for for any $\varepsilon>0$ any $\delta>0$ will be fine? $\endgroup$ – JaVaPG Aug 13 '14 at 23:34
  • $\begingroup$ Not quite. What you have shown is that for any $\epsilon > 0$, $\delta = \frac{1}{2}$ works. $\endgroup$ – JimmyK4542 Aug 13 '14 at 23:35
  • $\begingroup$ Other question if I may, In case $\delta=1$ the problem cannot be solved? right? $\endgroup$ – JaVaPG Aug 13 '14 at 23:39
  • $\begingroup$ You can't use $\delta = 1$ to solve the problem. But that doesn't matter. For any $\epsilon > 0$ you get to choose the $\delta > 0$. So for this problem, don't pick $\delta = 1$. $\endgroup$ – JimmyK4542 Aug 13 '14 at 23:40

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.