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Question Number 121817 by shaker last updated on 12/Nov/20

Answered by liberty last updated on 12/Nov/20

$$\underset{x\to 1}{\lim }\left(\frac{{\mathrm{nx}}^{\mathrm{m}}-\mathrm{n}-{\mathrm{mx}}^{\mathrm{n}}+\mathrm{m}}{{\mathrm{x}}^{\mathrm{m}+\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{m}}+1}\right)=$$$$\underset{x\to 1}{\lim }\left(\frac{\mathrm{mn}({\mathrm{x}}^{\mathrm{m}-1}-{\mathrm{x}}^{\mathrm{n}-1})}{(\mathrm{m}+\mathrm{n}){\mathrm{x}}^{\mathrm{m}+\mathrm{n}-1}-{\mathrm{nx}}^{\mathrm{n}-1}-{\mathrm{mx}}^{\mathrm{m}-1}}\right)=$$$$\underset{x\to 1}{\lim }\left(\frac{\mathrm{mn}((\mathrm{m}-1){\mathrm{x}}^{\mathrm{m}-2}-(\mathrm{n}-1){\mathrm{x}}^{\mathrm{n}-2})}{(\mathrm{m}+\mathrm{n})(\mathrm{m}+\mathrm{n}-1){\mathrm{x}}^{\mathrm{m}+\mathrm{n}-2}-\mathrm{n}(\mathrm{n}-1){\mathrm{x}}^{\mathrm{n}-2}-\mathrm{m}(\mathrm{m}-1){\mathrm{x}}^{\mathrm{m}-2}}\right)=$$$$\frac{\mathrm{mn}(\mathrm{m}-\mathrm{n})}{{\mathrm{m}}^2+2\mathrm{mn}+{\mathrm{n}}^2-\mathrm{m}-\mathrm{n}-{\mathrm{n}}^2+\mathrm{n}-{\mathrm{m}}^2+\mathrm{m}}$$$$=\frac{\mathrm{mn}(\mathrm{m}-\mathrm{n})}{2\mathrm{mn}}=\frac{\mathrm{m}-\mathrm{n}}{2}.▴$$

Answered by bemath last updated on 12/Nov/20

$$letx=1+w;w\to 0$$$$\underset{w\to 0}{\lim }(\frac{n}{{(1+w)}^n-1}-\frac{m}{{(1+w)}^m-1})$$$$=\underset{w\to 0}{\lim }(\frac{n}{nw+\frac{n(n-1)}{2}{w}^2}-\frac{m}{mw+\frac{m(m-1)}{2}{w}^2})$$$$=\underset{w\to 0}{\lim }(\frac{2n}{2nw+n(n-1)w^2}-\frac{2m}{2mw+m(m-1)w^2})$$$$=\underset{w\to 0}{\lim }\left(\frac{2n(2mw+m(m-1)w^2)-2m(2nw+n(n-1)w^2)}{(2nw+n(n-1)w^2)(2mw+m(m-1)w^2)}\right)$$$$=\underset{w\to 0}{\lim }\left(\frac{4mn+2mn(m-1)w-4mn-2mn(n-1)w}{(2n+n(n-1)w)(2mw+m(m-1)w^2)}\right)$$$$=\underset{w\to 0}{\lim }\left(\frac{(2mn(m-1)-2mn(n-1))w}{(2n+n(n-1)w)(2m+m(m-1)w)w}\right)$$$$=\underset{w\to 0}{\lim }(\frac{2mn(m-1-n+1)}{(2n+n(n-1)w)(2m+m(m-1)w)}$$$$=\underset{w\to 0}{\lim }\left(\frac{2mn.(m-n)}{2n.2m}\right)=\frac{m-n}{2}$$

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