91Ó°ÊÓ

Firstly, recognize that the Taylor series expansion for \(e^x\) around a point \(x=0\) is \[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] which is equal to the sum of \[e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\] Now, for all \(x>0\) we have \[e^x > 1+x \] because the terms \(\frac{x^2}{2!}, \frac{x^3}{3!},... \) are all positive for \(x>0\).

Now, using Lagrange's theorem, we know that for every function \(f(x)\) which is \(n+1\) times continuously differentiable on an interval \([a,x]\), for any \(x\) in \([a,x]\) there exists a \(\xi\) in \([a,x]\) such that \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + … + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) \) where \(R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{(n+1)} \) is the remainder term. For the function \( f(x) = e^x \), because \(e^x\) is infinitely differentiable and its derivative of any order is also \( e^x \) at any point \(x\), applying Lagrange's formula on \(e^x\) with \(n = 1\), we get \( e^x = 1 + x + \frac{e^\xi}{2!}x^2 \) where \( \xi \) is in \([0,x]\). Therefore, for \(x>0\), \(\xi > 0\) and we have \(e^x < 1 + x + \frac{e^x}{2}x^2 \). Dividing by the positive term \( \frac{e^x}{2} \), we simplify the above relation to \( e^x < 1 + x e^x \)

Combining the results from the two steps, for any \(x>0\), we have \(1+x < e^x < 1 + xe^x\).