In Maclaurin's expansion, show that if $f^{n+1}(x)\ne 0$ is continuous and $\theta_n$ is a number between $0$ and $1$ that appears in $R_n(x)=\frac{x^n}{n!}.f^{(n)}(\theta_nx)$ then $\theta_n\to \frac{1}{n+1}$ as $x\to0$.
What I did was, since $R_n(x)=\frac{x^{n+1}}{(n+1)!}.f^{(n+1)}(\phi x)$, where $0< \phi <1$, we get: $$\theta_n=\frac{f^{(n+1)}(\phi x)}{(n+1).f^{(n)}(\theta_nx)}.\theta_nx$$ Now, I'll have to prove that $$ \lim_{x\to 0}\frac{f^{(n+1)}(\phi x)}{f^{(n)}(\theta_nx)}.\theta_nx=1$$ But all I'm able to deduce is that $\lim_{x\to0}f^{(n)}(\theta_nx)=0$