Question

`\lim_(n\rarr\infty)(1-4/n)^n`: convergent or divergent?

I know the answer is "converges to `1/e^4`," but I have no idea how you get there :(

Answer 1

Yes, the limit does converge to `e^(-4)`.

Explanation:

Instead of solving this particular case, let's do the general one; let

`L = lim_(n->oo)(1+a/n)^(bn)`.

Now, if both `a` and `b` are equal to `1`, `L = e`.

`lim_(n->oo) (1+1/n)^n = e`.

This is one of the definitions of `e`.

Now, let `t = 1/n`. While `n->oo`, `t` much approach `0`. Therefore, we have

`lim_(n->oo)(1+1/n)^n = lim_(t->0)(1+t)^(1/t)`

This new limit is still equal to `e`.

`lim_(t->0)(1+t)^(1/t) = e`.

Let's do the same with our general case.

`L = lim_(n->oo)(1+a/n)^(bn)`

Let `t = a/n`. `t` still approaches `0`, as `a` is finite.

Because `t=a/n => n = a/t`.

Our limit has become

`L = lim_(t->0)(1+t)^((ab)/t) = (lim_(t->0)(1+t)^(1/t))^(ab)`

But we know that the inner limit is `e`, therefore

`color(red)(L = e^(ab)`

`lim_(n->oo)(1+a/n)^(bn) = e^(ab)`.

In our particular case, `a = -4` and `b=1`. By plugging these into the limit, you get:

`lim_(n->oo)(1+(-4)/n)^(1*n) = e^(-4*1) = 1/e^4`.