Question

How do you evaluate the limit `(1+1/sqrtx)^x` as x approaches `oo`?

Answer 1

`(1+1/sqrt(x))^x -> oo` as `x->oo`

Explanation:

`lim_(x->oo)(1+1/sqrt(x))^x = lim_(x->oo)e^(ln((1+1/sqrt(x))^x))`

`=lim_(x->oo)e^(xln(1+1/sqrt(x)))`

`=e^(lim_(x->oo)xln(1+1/sqrt(x)))" (*)"`

(The previous step is true as `e^x` is a continuous function)

`lim_(x->oo)xln(1+1/sqrt(x)) = lim_(x->oo)ln(1+1/sqrt(x))/(1/x)`

`=lim_(x->oo)(d/dxln(1+1/sqrt(x)))/(d/dx1/x)` (by L'Hopital's rule)

`=lim_(x->oo)(1/(1+1/sqrt(x))*1/(xsqrt(x)))/(1/x^2)`

`=lim_(x->oo)x/(sqrt(x)+1)`

`=oo`

Substituting this into `"(*)"`:

`lim_(x->oo)(1+1/sqrt(x))^x = e^oo = oo`

Thus `(1+1/sqrt(x))^x -> oo` as `x->oo`

Answer 2

`oo`

Explanation:

Another approach using Bernoulli's compounding formula ie

`\lim_{n to oo } (1+1/n)^{n} = e`

So here:
`lim_(x to oo) (1+1/sqrtx)^x`

`= lim_(x to oo) ( (1+1/sqrtx)^(sqrt x))^(sqrt x)`

with `u = sqrt x`

`= lim_(u to oo) ( (1+1/u)^u)^u`

`= lim_(u to oo) e^u = oo`

Answer 3

`oo`

Explanation:

Using the binomial expansion

`(1+u)^r = sum_(k=0)^oo ((r),(k))u^k`

making `y = sqrt(x)`

`z = (1+1/y)^(y^2) = 1+y^2(1/y)+(y^2(y^2-1))/(1 xx 2)(1/y)^2+cdots`

`z = 1 + y + f(y)`

`lim_(y->oo)z = oo` because `lim_(y->oo)f(y) ge 0`