Question Number 121237 by benjo_mathlover last updated on 06/Nov/20
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{mx}\right)^{\mathrm{n}} −\left(\mathrm{1}+\mathrm{nx}\right)^{\mathrm{m}} }{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$
Answered by liberty last updated on 06/Nov/20
![L′Hopital lim_(x→0) ((mn(1+mx)^(n−1) −mn(1+nx)^(m−1) )/(2x)) mn ×lim_(x→0) ((m(n−1)(1+mx)^(n−2) −n(m−1)(1+nx)^(m−2) )/2) ((mn)/2) × [ mn−m−mn+n ] = ((mn(n−m))/2)](https://www.tinkutara.com/question/Q121238.png)
$$\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{mn}\left(\mathrm{1}+\mathrm{mx}\right)^{\mathrm{n}−\mathrm{1}} −\mathrm{mn}\left(\mathrm{1}+\mathrm{nx}\right)^{\mathrm{m}−\mathrm{1}} }{\mathrm{2x}} \\ $$$$\:\:\mathrm{mn}\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{m}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{mx}\right)^{\mathrm{n}−\mathrm{2}} −\mathrm{n}\left(\mathrm{m}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{nx}\right)^{\mathrm{m}−\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\frac{\mathrm{mn}}{\mathrm{2}}\:×\:\left[\:\mathrm{mn}−\mathrm{m}−\mathrm{mn}+\mathrm{n}\:\right] \\ $$$$=\:\frac{\mathrm{mn}\left(\mathrm{n}−\mathrm{m}\right)}{\mathrm{2}} \\ $$
Answered by Dwaipayan Shikari last updated on 06/Nov/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+{mnx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{m}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{1}−{mnx}−\frac{{m}\left({m}−\mathrm{1}\right)}{\mathrm{2}!}{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$={x}^{\mathrm{2}} \frac{{n}\left({n}−\mathrm{1}\right){m}^{\mathrm{2}} −{n}^{\mathrm{2}} {m}\left({m}−\mathrm{1}\right)}{{x}^{\mathrm{2}} \mathrm{2}!}=\frac{−{nm}^{\mathrm{2}} +{n}^{\mathrm{2}} {m}}{\mathrm{2}} \\ $$$$=\frac{{mn}\left({n}−{m}\right)}{\mathrm{2}} \\ $$