Question

Find the limit lim x→0^+ (1+x^2)^1/x?

find the lim`lim x→0^+ (1+x^2)^(1/x)`

Answer 1

`1`

Explanation:

let `y=lim_(xrarr0^+)(1+x^2)^(1/x)`
`ln(y)=lim_(xrarr0^+)ln((1+x^2)^(1/x))`
`ln(y)=lim_(xrarr0^+)ln(1+x^2)/x`

plugging `x=0` directly makes `0/0`, so use l'hopital's rule

`ln(y)=lim_(xrarr0^+)(d/dx(ln(1+x^2)))/(d/dx(x))`

`ln(y)=lim_(xrarr0^+)((2x)/(1+x^2))/1`

plugging `x=0` directly makes `0/1=0`

`ln(y)=0`
`y=e^0=1=lim_(xrarr0^+)(1+x^2)^(1/x)`

you can see `y` approaches 1 as `x` approaches 0
graph{(1+x^2)^(1/x) [-7.024, 7.024, -3.51, 3.513]}