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

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

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

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`