# Show that : lim(1+1/x)^x=e as x approaches infinity while taking on positive or negative real values.

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### 1 Answer

It has to be shown that `lim_(x->oo)(1+1/x)^x = e`

`lim_(x->oo)(1 + 1/x)^x = lim_(x->0)(1 + x)^(1/x)`

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

` `take the natural log of both the sides

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

=> `ln y = (1/x)*ln(1+x)`

As x tends to 0,

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

For x = 0, `(ln(1+x))/x = 0/0` which is indeterminate. This allows the use of the l'Hopital's rule and we can replace the numerator and denominator by their derivatives.

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

substituting x = 0, we get `ln y = 1`

=> y = e^1

=> `(1+x)^(1/x) = e`

**This shows that `lim_(x->0)(1+x)^(1/x) = e` or `lim_(x->oo)(1+1/x)^x = e` **