Question

How do you find the limit of `(e^x + x)^(1/x)` as x approaches 0?

Answer 1

`lim_{x\to0}h(x)=e^2`

Explanation:

Given `h(x)=(e^x+x)^(1/x)`

First, we'll check how the base comes out if we apply the limit `x\to0`. It seems that if we do so, then `e^x+x=e^0+0=1`
But, check the power, it seems that then the equation would become `1^\infty`. We could leave it at that but since this is limits, we need to be careful here.

So, to solve this we have an equation, that is if `h(x)=g(x)^f(x)` and as `x\toa` such that `g(x)\to0` and `f(x)\toinfty`, then we have a solution which is
`lim_{x\toa}h(x)=e^(h(x)*(g(x)-1)`

So, substituting `g(x)=e^x+x` and `f(x)=1/x`, we get
`lim_{x\to0}h(x)=e^(1/x(e^x+x-1)`

Let's now just take care of the powers, so we get
`lim_{x\to0}(e^x+x-1)/x=lim_{h\to0}(e^x-1)/x+x/x`
We know about the `(e^x-1)/x` at `x` tends to `0` limit equaling `1`, so we get it in the end as `1+1=2`

So, that is why there's an `e` in the answer.