# Calculus - prove limit using esplion-delta

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 :

$$0<\frac{1}{x}<1$$

Therefore :

$$\left|\left[\frac{1}{x}\right]-1\right|=|0-1|=1<\varepsilon$$

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 $1-\delta < x < 1$.
• 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? – JaVaPG Aug 13 '14 at 23:34
• Not quite. What you have shown is that for any $\epsilon > 0$, $\delta = \frac{1}{2}$ works. – JimmyK4542 Aug 13 '14 at 23:35
• Other question if I may, In case $\delta=1$ the problem cannot be solved? right? – JaVaPG Aug 13 '14 at 23:39
• 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 $\delta = 1$. – JimmyK4542 Aug 13 '14 at 23:40