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 $1-\delta < x < 1$.