Is my approach to this limit true?

Question

$$\lim_{n\to\infty}\lim_{t\to\infty}\bigg(\frac{t-n}{t}\bigg)$$

Using the limit definition of $e$

$$\lim_{n\to\infty}\bigg(1+\frac{1}{n}\bigg)^n=e$$

the required limit can be written as

$$\lim_{n\to\infty}\lim_{t\to\infty}\bigg(1-\frac{n}{t}\bigg)$$ $$\lim_{n\to\infty}\lim_{t\to\infty}\bigg(1-\frac{n}{t}\bigg)^{t/t}$$ $$\lim_{n\to\infty}\lim_{t\to\infty} e^{-n/t}=\frac{1}{e}$$

Is my approach true?

I have solved this limit using the limit definition of e. But the answer I am getting is completely different from the original answer i.e 1. Please help me in identifying the correct answer.

Answer 1

As pointed out in comments, directly take the limit first on $t$ to get $\displaystyle \lim_{n \to \infty} 1 =1$.

Secondly, $\displaystyle \lim_{t \to \infty}e^{-n/t}=e^0=1$.

Answer 2

You write

$$\lim_{n\to\infty}\lim_{t\to\infty} e^{-n/t}=\frac1e$$

but this is not justified because it is said nowhere that the ratio $\dfrac nt$ tends to $1$.

The correct way is

$$\lim_{n\to\infty}\left(\lim_{t\to\infty} e^{-n/t}\right)=\lim_{n\to\infty}1=1.$$