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`