Question

`lim_(x->0^+)((x+1)^x + 1)^x=`?

Answer 1

`1`

Explanation:

This limit may seem tricky, but it actually leads to no undefined forms (like `0^0` for example).

Let's see what happens as `x \to 0^+`: is we substitute all `x`'s with zeroes, the expression is approaching

`(\color(blue)((x+1)^x)\color(red)(+1))^x \to (\color(blue)((0+1)^0)\color(red)(+1))^0 \to (\color(blue)(1^0)\color(red)(+1))^0 \to (\color(blue)(1)\color(red)(+1))^0`

So, while the content of the parenthesis approaches `2`, its exponent approaches zero. The limit behaves like `2^x`, with `x \to 0`, and thus approaches `2^0=1`