lim-x-0-1-mx-n-1-nx-m-x-2-

Question Number 121237 by benjo_mathlover last updated on 06/Nov/20

\lim_{x \to 0} \frac{{(1 + mx)}^{n} - {(1 + nx)}^{m}}{x^{2}} = ?

Answered by liberty last updated on 06/Nov/20

L^{'} Hopital

\lim_{x \to 0} \frac{mn {(1 + mx)}^{n - 1} - mn {(1 + nx)}^{m - 1}}{2 x}

mn \times \lim_{x \to 0} \frac{m (n - 1) {(1 + mx)}^{n - 2} - n (m - 1) {(1 + nx)}^{m - 2}}{2}

\frac{mn}{2} \times [mn - m - mn + n]

= \frac{mn (n - m)}{2}

Answered by Dwaipayan Shikari last updated on 06/Nov/20

\lim_{x \to 0} \frac{1 + m n x + \frac{n (n - 1)}{2!} m^{2} x^{2} - 1 - m n x - \frac{m (m - 1)}{2!} n^{2} x^{2}}{x^{2}}

= x^{2} \frac{n (n - 1) m^{2} - n^{2} m (m - 1)}{x^{2} 2!} = \frac{- {n m}^{2} + n^{2} m}{2}

= \frac{m n (n - m)}{2}