$A=\int_\lambda^\infty \frac{x}{\lambda}e^{-x^2/2}dx, B=\int_\lambda^\infty e^{-x^2/2}dx$, calculate $\lim_{\lambda\to \infty}A/B$. $$A=\int_\lambda^\infty \frac{x}{\lambda}e^{-x^2/2}dx$$ $$B=\int_\lambda^\infty e^{-x^2/2}dx$$ I want to find: $$\boxed{\lim_{\lambda\to \infty}\frac AB}$$
I think the limit is $1$. But I have some confusions. If I use the L'hopital's rule, can I differentiate w.r.t $\lambda$? I am not sure how to deal with the indefinite integral? Can I treat the $\infty$ as constant？
Could someone kindly help? Thanks!

$$A(\lambda)=\frac1\lambda\int_{\lambda}^\infty x e^{-x^2/2} dx=\frac1\lambda e^{-\frac12\lambda^2}\\
A'(\lambda)=-\frac{1}{\lambda^2}e^{-\frac12\lambda^2}(\lambda^2+1)$$
$$B(\lambda)=\int_{\lambda}^\infty e^{-\frac12x^2}dx\\
B'(\lambda)=-e^{-\frac12\lambda^2}$$
$$\lim_{\lambda\to\infty}\frac{A(\lambda)}{B(\lambda)}=\lim_{\lambda\to\infty}\frac{A'(\lambda)}{B'(\lambda)}=\lim_{\lambda\to\infty}\frac{(\lambda^2+1)}{\lambda^2}\frac{e^{-\frac12\lambda^2}}{e^{-\frac12\lambda^2}}=1$$
