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Question Number 160077 by qaz last updated on 24/Nov/21

lim_(x→1) (((654)/(1−x^(654) ))−((678)/(1−x^(678) )))=?

$$\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{654}}{\mathrm{1}−\mathrm{x}^{\mathrm{654}} }−\frac{\mathrm{678}}{\mathrm{1}−\mathrm{x}^{\mathrm{678}} }\right)=? \\ $$

Commented by cortano last updated on 25/Nov/21

lim_(x→1) (((654)/(1−x^(654) )) − ((678)/(1−x^(678) ))) let 1−x^(654) = u ; x^(654) =1−u x^(678) =x^(654) .x^(24) =(1−u)(1−u)^((24)/(654)) = lim_(u→0) (((654)/u) −((678)/(1−(1−u)^((113)/(109)) )) ) = lim_(u→0) (((654(1−(1−u)^((113)/(109)) )−678u)/(u(1−(1−u)^((113)/(109)) )))) = lim_(u→0) (((654(1−(1−((113)/(109))u+((226)/(11881))u^2 )−678u)/(u(1−(1−((113)/(109))u+((226)/(11881))u^2 ))))) =lim_(u→0) (((654(((113)/(109))u−((226)/(11881))u^2 )−678u)/(u(1−(1−((113)/(109))u+((226)/(11881))u^2 ))))) = lim_(u→0) (((−((1356)/(109))u^2 )/(u^2 (((113)/(109))−((226)/(11881))u)))) = −((1356)/(113))=−12

$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{654}}{\mathrm{1}−{x}^{\mathrm{654}} }\:−\:\frac{\mathrm{678}}{\mathrm{1}−{x}^{\mathrm{678}} }\right) \\ $$$$\:{let}\:\mathrm{1}−{x}^{\mathrm{654}} \:=\:{u}\:;\:{x}^{\mathrm{654}} =\mathrm{1}−{u} \\ $$$$\:{x}^{\mathrm{678}} ={x}^{\mathrm{654}} .{x}^{\mathrm{24}} =\left(\mathrm{1}−{u}\right)\left(\mathrm{1}−{u}\right)^{\frac{\mathrm{24}}{\mathrm{654}}} \\ $$$$=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{654}}{{u}}\:−\frac{\mathrm{678}}{\mathrm{1}−\left(\mathrm{1}−{u}\right)^{\frac{\mathrm{113}}{\mathrm{109}}} }\:\right) \\ $$$$=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{654}\left(\mathrm{1}−\left(\mathrm{1}−{u}\right)^{\frac{\mathrm{113}}{\mathrm{109}}} \right)−\mathrm{678}{u}}{{u}\left(\mathrm{1}−\left(\mathrm{1}−{u}\right)^{\frac{\mathrm{113}}{\mathrm{109}}} \right)}\right) \\ $$$$=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{654}\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{113}}{\mathrm{109}}{u}+\frac{\mathrm{226}}{\mathrm{11881}}{u}^{\mathrm{2}} \right)−\mathrm{678}{u}\right.}{{u}\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{113}}{\mathrm{109}}{u}+\frac{\mathrm{226}}{\mathrm{11881}}{u}^{\mathrm{2}} \right)\right)}\right) \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{654}\left(\frac{\mathrm{113}}{\mathrm{109}}{u}−\frac{\mathrm{226}}{\mathrm{11881}}{u}^{\mathrm{2}} \right)−\mathrm{678}{u}}{{u}\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{113}}{\mathrm{109}}{u}+\frac{\mathrm{226}}{\mathrm{11881}}{u}^{\mathrm{2}} \right)\right)}\right) \\ $$$$=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{−\frac{\mathrm{1356}}{\mathrm{109}}{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} \left(\frac{\mathrm{113}}{\mathrm{109}}−\frac{\mathrm{226}}{\mathrm{11881}}{u}\right)}\right) \\ $$$$\:=\:−\frac{\mathrm{1356}}{\mathrm{113}}=−\mathrm{12} \\ $$

Commented by blackmamba last updated on 24/Nov/21

correct

$${correct} \\ $$

Commented by qaz last updated on 24/Nov/21

thanks a lot

$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\ $$

Commented by bobhans last updated on 25/Nov/21

lim_(x→1) (((654(1−x^(678) )−678(1−x^(654) ))/((1−x^(654) )(1−x^(678) )))) = lim_(x→1) (((−24−654x^(678) +678x^(654) )/(1−x^(654) −x^(678) +x^(1332) ))) =lim_(x→1) (((−443412x^(677) +443412x^(653) )/(−654x^(653) −678x^(677) +1332x^(1331) ))) = lim_(x→1) (((−300189924x^(676) +289548036x^(652) )/(−427062x^(652) −459006x^(676) +1772892x^(1330) ))) =((−10641888)/(886824)) =−12

$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{654}\left(\mathrm{1}−\mathrm{x}^{\mathrm{678}} \right)−\mathrm{678}\left(\mathrm{1}−\mathrm{x}^{\mathrm{654}} \right)}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{654}} \right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{678}} \right)}\right) \\ $$$$=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{−\mathrm{24}−\mathrm{654x}^{\mathrm{678}} +\mathrm{678x}^{\mathrm{654}} }{\mathrm{1}−\mathrm{x}^{\mathrm{654}} −\mathrm{x}^{\mathrm{678}} +\mathrm{x}^{\mathrm{1332}} }\right) \\ $$$$\:=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{−\mathrm{443412x}^{\mathrm{677}} +\mathrm{443412x}^{\mathrm{653}} }{−\mathrm{654x}^{\mathrm{653}} −\mathrm{678x}^{\mathrm{677}} +\mathrm{1332x}^{\mathrm{1331}} }\right) \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{−\mathrm{300189924x}^{\mathrm{676}} +\mathrm{289548036x}^{\mathrm{652}} }{−\mathrm{427062x}^{\mathrm{652}} −\mathrm{459006x}^{\mathrm{676}} +\mathrm{1772892x}^{\mathrm{1330}} }\right) \\ $$$$\:=\frac{−\mathrm{10641888}}{\mathrm{886824}}\:=−\mathrm{12} \\ $$

Answered by FongXD last updated on 24/Nov/21

L=lim_(x→1) ((m/(1−x^m ))−(n/(1−x^n ))) L=lim_(x→1) ((m(1−x^n )−n(1−x^m ))/((1−x^m )(1−x^n ))) L=lim_(x→1) (((1−x)[m(1+x+...+x^(n−1) )−n(1+x+...+x^(m−1) )])/((1−x)(1+x+...+x^(m−1) )(1−x)(1+x+...+x^(n−1) ))) L=lim_(x→1) ((m(x^(n−1) +...+x+1)−m(1+...+1+1^(n times) )−n(x^(m−1) +...+x+1)+n(1+...+1+1^(m times) ))/((1−x)(1+x+...+x^(m−1) )(1+x+...+x^(n−1) ))) L=lim_(x→1) ((m[(x^(n−1) −1)+...+(x^2 −1)+(x−1)]−n[(x^(m−1) −1)+...+(x^2 −1)+(x−1)])/((1−x)(1+x+...+x^(m−1) )(1+x+...+x^(n−1) ))) L=lim_(x→1) (((x−1){m[(x^(n−2) +...+x+1)+...+(x+1)+1]−n[(x^(m−2) +...+x+1)+...+(x+1)+1]})/((1−x)(1+x+...+x^(m−1) )(1+x+...+x^(n−1) ))) L=((m[(1+...+1+1^(n−1 times) )+...+(1+1)+1]−n[(1+...+1+1^(m−1 times) )+...+(1+1)+1])/(−(1+1+...+1_(m times) )(1+1+...+1_(n times) ))) L=((m(1+2+...+(n−1))−n(1+2+...+(m−1)))/(−mn)) L=((((mn(n−1))/2)−((nm(m−1))/2))/(−mn))=((m−n)/2) take m=654 and n=678, ⇒ L=((654−678)/2)=−12

$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{m}}{\mathrm{1}−\mathrm{x}^{\mathrm{m}} }−\frac{\mathrm{n}}{\mathrm{1}−\mathrm{x}^{\mathrm{n}} }\right) \\ $$$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{m}\left(\mathrm{1}−\mathrm{x}^{\mathrm{n}} \right)−\mathrm{n}\left(\mathrm{1}−\mathrm{x}^{\mathrm{m}} \right)}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{m}} \right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{n}} \right)} \\ $$$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left(\mathrm{1}−\mathrm{x}\right)\left[\mathrm{m}\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{n}−\mathrm{1}} \right)−\mathrm{n}\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{m}−\mathrm{1}} \right)\right]}{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{m}−\mathrm{1}} \right)\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{n}−\mathrm{1}} \right)} \\ $$$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{m}\left(\mathrm{x}^{\mathrm{n}−\mathrm{1}} +...+\mathrm{x}+\mathrm{1}\right)−\mathrm{m}\left(\overset{\mathrm{n}\:\mathrm{times}} {\mathrm{1}+...+\mathrm{1}+\mathrm{1}}\right)−\mathrm{n}\left(\mathrm{x}^{\mathrm{m}−\mathrm{1}} +...+\mathrm{x}+\mathrm{1}\right)+\mathrm{n}\left(\overset{\mathrm{m}\:\mathrm{times}} {\mathrm{1}+...+\mathrm{1}+\mathrm{1}}\right)}{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{m}−\mathrm{1}} \right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{n}−\mathrm{1}} \right)} \\ $$$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{m}\left[\left(\mathrm{x}^{\mathrm{n}−\mathrm{1}} −\mathrm{1}\right)+...+\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)+\left(\mathrm{x}−\mathrm{1}\right)\right]−\mathrm{n}\left[\left(\mathrm{x}^{\mathrm{m}−\mathrm{1}} −\mathrm{1}\right)+...+\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)+\left(\mathrm{x}−\mathrm{1}\right)\right]}{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{m}−\mathrm{1}} \right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{n}−\mathrm{1}} \right)} \\ $$$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left(\mathrm{x}−\mathrm{1}\right)\left\{\mathrm{m}\left[\left(\mathrm{x}^{\mathrm{n}−\mathrm{2}} +...+\mathrm{x}+\mathrm{1}\right)+...+\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{1}\right]−\mathrm{n}\left[\left(\mathrm{x}^{\mathrm{m}−\mathrm{2}} +...+\mathrm{x}+\mathrm{1}\right)+...+\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{1}\right]\right\}}{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{m}−\mathrm{1}} \right)\left(\mathrm{1}+\mathrm{x}+...+\mathrm{x}^{\mathrm{n}−\mathrm{1}} \right)} \\ $$$$\mathrm{L}=\frac{\mathrm{m}\left[\left(\overset{\mathrm{n}−\mathrm{1}\:\mathrm{times}} {\mathrm{1}+...+\mathrm{1}+\mathrm{1}}\right)+...+\left(\mathrm{1}+\mathrm{1}\right)+\mathrm{1}\right]−\mathrm{n}\left[\left(\overset{\mathrm{m}−\mathrm{1}\:\mathrm{times}} {\mathrm{1}+...+\mathrm{1}+\mathrm{1}}\right)+...+\left(\mathrm{1}+\mathrm{1}\right)+\mathrm{1}\right]}{−\left(\underset{\mathrm{m}\:\mathrm{times}} {\mathrm{1}+\mathrm{1}+...+\mathrm{1}}\right)\left(\underset{\mathrm{n}\:\mathrm{times}} {\mathrm{1}+\mathrm{1}+...+\mathrm{1}}\right)} \\ $$$$\mathrm{L}=\frac{\mathrm{m}\left(\mathrm{1}+\mathrm{2}+...+\left(\mathrm{n}−\mathrm{1}\right)\right)−\mathrm{n}\left(\mathrm{1}+\mathrm{2}+...+\left(\mathrm{m}−\mathrm{1}\right)\right)}{−\mathrm{mn}} \\ $$$$\mathrm{L}=\frac{\frac{\mathrm{mn}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{nm}\left(\mathrm{m}−\mathrm{1}\right)}{\mathrm{2}}}{−\mathrm{mn}}=\frac{\mathrm{m}−\mathrm{n}}{\mathrm{2}} \\ $$$$\mathrm{take}\:\mathrm{m}=\mathrm{654}\:\mathrm{and}\:\mathrm{n}=\mathrm{678},\:\Rightarrow\:\mathrm{L}=\frac{\mathrm{654}−\mathrm{678}}{\mathrm{2}}=−\mathrm{12} \\ $$

Commented by mr W last updated on 24/Nov/21

great!

$${great}! \\ $$

Commented by FongXD last updated on 24/Nov/21

thanks sir!

$$\mathrm{thanks}\:\mathrm{sir}! \\ $$

Commented by qaz last updated on 24/Nov/21

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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