If $x > 0$ and $t \leq x $
Prove:
$ e^{-t}-\left(1-\dfrac{t}{x}\right)^x \geq 0 \>. $
If $x > 0$ and $t \leq x $
Prove:
$ e^{-t}-\left(1-\dfrac{t}{x}\right)^x \geq 0 \>. $
This is equivalent to $\mathrm e^{-t}\geqslant\left(1-t/x\right)^x$, which is equivalent to $\mathrm e^{-u}\geqslant1-u$ for $u=t/x$.
Now, $u\mapsto1-u$ is the tangent to the graph of the function $u\mapsto\mathrm e^{-u}$ at $u=0$. This function is convex hence its graph lies above any of its tangents and you are done.
Or, consider the function $b:u\mapsto\mathrm e^{-u}-1+u$ and note that $b(0)=0$ and b'(u)=1-\mathrm e^{-u}. Hence the function $b$ is decreasing on $u\leqslant0$ and increasing on $u\geqslant0$. This proves that $b\geqslant0$ everywhere.