# What is `lim_(y->oo)(1+1/y)^y`

*print*Print*list*Cite

### 1 Answer

The limit: `lim_(y->oo)(1+1/y)^y`

As ` `y tends to infinity, 1/y tends to 0. If 1/y is denoted as x, the given limit can be written as `lim_(x->0)(1 + x)^(1/x)`

Let `z = (1+x)^(1/x)`

Take the natural logarithm of both the sides.

`ln z = ln(1+x)^(1/x)`

=> `ln z = (ln(1+x))/x`

for x tending to 0

`ln z = lim_(x->0)(ln(1+x))/x`

substituting x=0, gives an indeterminate form, this allows the use of l'Hopital's rule and the numerator and denominator of the limit can be substituted by their derivatives.

`ln z = lim_(x-> 0)(1/(1+x))`

substituting x = 0

=> `ln z = 1`

=> z = e^1

**The required limit `lim_(y->oo)(1+1/y)^y = e` **