If $f(x)=-e^{-x}$ on $(0,\infty)$, Prove that $inf_{x>0}$ $f(x)=-1$.
For all $x>0$, $f(x)>-1$ then $-1$ is a lower bound of $f$ on $(0,\infty)$.
I did it up to here. And I want to find a any $\delta$ to solve the problem, as in the following example. But I don't know what to do. I need some help.
Example: $g(x)=\frac{1}{x}$, Prove $\inf_{x>0}$ $g(x)=0$.
Since $0<g(x)$ for every $x>0$, $0$ is a lower bound of f on $(0,\infty)$. Suppose $\delta$ is any positive real number. If we choose $x>\frac{1}{\delta}$, then $f(x)=\frac{1}{x}<\delta$ and so $\delta$ is not a lower bound of f on $(0,\infty)$. Since $\delta>0$ was arbitrary, if follows that $0=\inf_{x>0}f(x)$.
$f(x)=\frac{-1}{e^x}$
$f'(x)=\frac{1}{e^x}$
Now, for any x in $(0,\infty)$, $f'(x)>0$,hence f is monotonically increasing function Hence, $\lim x\rightarrow0+$, f(x)=-1 is infimum of f on (0,$\infty$).
I think ,this one is very easy to observe.
Now, to prove it in your way, Assume, $ -1+\delta$ be inf of f on (0,$\infty$). There for some y in (0,$\infty$), $f(y)=-1+\delta$
Now, $\frac{-1}{e^y} =-1+\delta$
$e^y= \frac{1}{(1-\delta)}$
$y=\ln\frac{1}{(1-\delta)}$
choose some x <y from here On (0,$\infty$) & get answer