1
$\begingroup$

Calculate $$\lim_{n\to \infty} \bigg(1+\frac{\theta_n(x) x}{n}\bigg)^n$$ where $\theta_n$ is continuous on $\mathbb{R}$ and $\lim_{n\to \infty}\theta_n=0$ and we have a fixed $x\in \mathbb{R}$.

Here $\theta_n(x):=\theta(x/n)$ where $\theta$ is continuous at $0$ such that $\theta(0)=0$. I came across to this limit while proving a result from Probability. Usually I would solve such a limit by $\lim_{n\to \infty} \bigg(\bigg(1+\frac{\theta_n x}{n}\bigg)^{n/\theta_n}\bigg)^{\theta_n}$ and noting that the inside is $e$ so we get $e^{\theta_n}$ which converges to $1$. But I am wondering how to show this rigorously. I've always resorted to such a technique in solving these limits, but I would like to know how to achieve this rigorously. I would greatly appreciate any help.

  • 0
    Do you mean $\theta_n(x)$? The way you have written it makes $\theta_n$ out to both be and not be a function of something. – 2017-02-09
  • 0
    Perhaps $\theta(x)=\theta(n)$ otherwise it's a family of real continuous functions. – 2017-02-09
  • 0
    Sorry I edited it to make it clearer. – 2017-02-09
  • 0
    My answer here may help: http://math.stackexchange.com/questions/374747/if-z-n-to-z-then-1z-n-nn-to-ez/374765#374765 – 2017-02-09
  • 0
    If $\theta_n(x)=0,$ then you have divided by $0$ in the power. – 2017-02-09

0 Answers 0