If-x-is-nearly-equal-to-1-then-mx-m-nx-n-m-n-

Question Number 44397 by Necxx last updated on 28/Sep/18

$${If}\:{x}\:{is}\:{nearly}\:{equal}\:{to}\:\mathrm{1}\:{then} \\ $$$$\frac{{mx}^{{m}} −{nx}^{{n}} }{{m}−{n}}= \\ $$

Commented by MrW3 last updated on 28/Sep/18

if x≈1 ⇒x^m ≈x^n ≈1 ((mx^m −nx^n )/(m−n))≈((m−n)/(m−n))=1

$${if}\:{x}\approx\mathrm{1} \\ $$$$\Rightarrow{x}^{{m}} \approx{x}^{{n}} \approx\mathrm{1} \\ $$$$\frac{{mx}^{{m}} −{nx}^{{n}} }{{m}−{n}}\approx\frac{{m}−{n}}{{m}−{n}}=\mathrm{1} \\ $$

Commented by Necxx last updated on 28/Sep/18

The options are a) x^(m+n) b)x^(m−n) c)x^m d)x^n I also got 1 but couldnt see it in option

$$\left.{T}\left.{he}\:{options}\:{are}\:{a}\right)\:{x}^{{m}+{n}} \:{b}\right){x}^{{m}−{n}} \\ $$$$\left.{c}\left.\right){x}^{{m}} \:{d}\right){x}^{{n}} \\ $$$${I}\:{also}\:{got}\:\mathrm{1}\:{but}\:{couldnt}\:{see}\:{it}\:{in} \\ $$$${option} \\ $$$$ \\ $$$$ \\ $$

Commented by MrW3 last updated on 28/Sep/18

$${all}\:{these}\:{options}\:{are}\:{also}\:\approx\mathrm{1} \\ $$

Commented by math khazana by abdo last updated on 29/Sep/18

changement x =1+ξ give ((mx^m −nx^n )/(m−n)) =((m(1+ξ)^m −n(1+ξ)^n )/(m−n)) =A(x) x→1 ⇒ξ→0 so (1+ξ)^m ∼1+mξ and (1+ξ)^n ∼1+nξ ⇒A(x) ∼ ((m(1+mξ)−n(1+nξ))/(m−n)) = 1 +(m+n)ξ →1 (ξ→0) ⇒ lim_(x→1) A(x)) =1 .

$${changement}\:{x}\:=\mathrm{1}+\xi\:{give}\: \\ $$$$\frac{{mx}^{{m}} −{nx}^{{n}} }{{m}−{n}}\:=\frac{{m}\left(\mathrm{1}+\xi\right)^{{m}} −{n}\left(\mathrm{1}+\xi\right)^{{n}} }{{m}−{n}}\:={A}\left({x}\right) \\ $$$${x}\rightarrow\mathrm{1}\:\Rightarrow\xi\rightarrow\mathrm{0}\:{so}\:\left(\mathrm{1}+\xi\right)^{{m}} \:\sim\mathrm{1}+{m}\xi\:{and} \\ $$$$\left(\mathrm{1}+\xi\right)^{{n}} \:\sim\mathrm{1}+{n}\xi\:\Rightarrow{A}\left({x}\right)\:\sim\:\frac{{m}\left(\mathrm{1}+{m}\xi\right)−{n}\left(\mathrm{1}+{n}\xi\right)}{{m}−{n}} \\ $$$$=\:\mathrm{1}\:+\left({m}+{n}\right)\xi\:\rightarrow\mathrm{1}\:\left(\xi\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\left.{lim}_{{x}\rightarrow\mathrm{1}} {A}\left({x}\right)\right)\:=\mathrm{1}\:. \\ $$

Answered by ajfour last updated on 28/Sep/18

$$\:=\:\frac{{m}\left(\mathrm{1}+{mh}\right)−{n}\left(\mathrm{1}+{nh}\right)}{{m}−{n}} \\ $$$$\:=\:\mathrm{1}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18

((mx^m −nx^m +nx^m −nx^n )/(m−n)) =((x^m (m−n)+nx^n (x^(m−n) −1))/(m−n)) =x^m +((x^n (x^(m−n) −1))/((m/n)−1)) x^(m−n) ≈1 =x^m

$$\frac{{mx}^{{m}} −{nx}^{{m}} +{nx}^{{m}} −{nx}^{{n}} }{{m}−{n}} \\ $$$$=\frac{{x}^{{m}} \left({m}−{n}\right)+{nx}^{{n}} \left({x}^{{m}−{n}} −\mathrm{1}\right)}{{m}−{n}} \\ $$$$={x}^{{m}} +\frac{{x}^{{n}} \left({x}^{{m}−{n}} −\mathrm{1}\right)}{\frac{{m}}{{n}}−\mathrm{1}} \\ $$$${x}^{{m}−{n}} \approx\mathrm{1} \\ $$$$={x}^{{m}} \\ $$

Commented by Necxx last updated on 28/Sep/18

$${Thats}\:{true}…\:{I}'{m}\:{clear}\:{now} \\ $$

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