Show that $1-\lambda h e^{-\lambda h} - e^{-\lambda h} = o(h)$

Question

We want to show that : $1-\lambda h e^{-\lambda h} - e^{-\lambda h} = o(h)$ with $\lambda>0$ a constant.

First, using the Taylor formula I have written this as $-o(1) - \lambda h - o(h)$ :

I use the fact that $e^{-\lambda h} = 1 + o(1)$ and $\lambda h e^{-\lambda h} = \lambda h + o(h)$

So $1-\lambda he^{-\lambda h} -e^{-\lambda h} = -o(1) -\lambda h - o(h) $ if I am right

So know I want to show that $\lim_{x\to a}\frac{-o(1) - \lambda h - o(h)}{h} = 0$

But I don't know how to prove this, in fact I don't know how to start, $o(h)$ is easy to handle but $\lambda h$ and $o(1)$ put me in trouble. For $\lambda h$ I don't see how can I obtain a $o(h)$ since lambda is strictly positive and constant

Does anyone have any tips on how to do this?

Thank you a lot

EDIT : From the comment of jjagmath so I thank him for that, hoping to have made this post clearer.

Answer 1

An answer using the $o$ notation:

Since $e^x = 1+x+o(x)$ when $x \to 0$ we have

\begin{align} 1-\lambda\ h\ e^{-\lambda\ h} - e^{-\lambda\ h} & = 1-\lambda\ h\ (1-\lambda\ h + o(h))-(1-\lambda\ h + o(h))\\ &=1-\lambda\ h+\lambda^2h^2+o(h^2)-1+\lambda\ h + o(h)\\ &=\lambda^2h^2+o(h^2) + o(h)\\ &=o(h)+o(h)+o(h) = o(h) \end{align}

when $h \to 0$.

Note that we used some properties of the $o$ notation as $o(\lambda\ h) = o(h)$, $o(h^2) = o(h)$ and $o(h) + o(h) = o(h)$.

Answer 2

What you need to prove is that $$\lim_{h\to\color{red}0}\frac{1-\lambda h e^{-\lambda h} - e^{-\lambda h}}{h} =0$$ which follows from L'Hôpital's rule.