Question

What is the value of `lim_(x->0)[(1+x)^(1/x)-e+1/2ex]/x^2` ?

Answer 1

`lim_(xrarr0)((1+x)^(1/x)-e+1/2ex)/x^2=oo`

Explanation:

First note that

`lim_(xrarr0)(1+x)^(1/x)=e`

This is actually how `e` was "invented". Here's a quick proof.

Let `L=lim_(xrarr0)(1+x)^(1/x)`.

Now take the natural log of both sides.

`lnL=ln[lim_(xrarr0)(1+x)^(1/x)]=lim_(xrarr0)ln(1+x)^(1/x)=lim_(xrarr0)ln(1+x)/x`

Now we use L'Hopital's Rule

`lnL=lim_(xrarr0)(1/(1+x))/1=1`

Therefore `L=lim_(xrarr0)(1+x)^(1/x)=e`.

So

`lim_(xrarr0)((1+x)^(1/x)-e+1/2ex)/x^2=lim_(xrarr0)(e-e+1/2ex)/x^2`

`=lim_(xrarr0)(e)/(2x)=oo`