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Question Number 184790 by Spillover last updated on 11/Jan/23
Evaluate lim_(x→0) (((1+x)^6 −1)/((1+x)^5 −1))
$${Evaluate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{6}} −\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{5}} −\mathrm{1}} \\ $$
Commented by MJS_new last updated on 11/Jan/23
l′Ho^� pital ⇒ (6/5)
$$\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\:\Rightarrow\:\frac{\mathrm{6}}{\mathrm{5}} \\ $$
Commented by Spillover last updated on 11/Jan/23
your right
$${your}\:{right} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Jan/23
Let 1+x=y, x→0⇒y→1 lim_(y→1) ((y^6 −1)/(y^5 −1))=lim_(y→1) (((y−1)(y+1)(y^2 −y+1)(y^2 +y+1))/((y−1)(y^4 +y^3 +y^2 +y+1))) =lim_(y→1) (((y+1)(y^2 −y+1)(y^2 +y+1))/((y^4 +y^3 +y^2 +y+1))) =(((1+1)(1^2 −1+1)(1^2 +1+1))/(1^4 +1^3 +1^2 +1+1))=(6/5)
$${Let}\:\mathrm{1}+{x}={y}, \\ $$$${x}\rightarrow\mathrm{0}\Rightarrow{y}\rightarrow\mathrm{1} \\ $$$$\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{y}^{\mathrm{6}} −\mathrm{1}}{{y}^{\mathrm{5}} −\mathrm{1}}=\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\cancel{\left({y}−\mathrm{1}\right)}\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} −{y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)}{\cancel{\left({y}−\mathrm{1}\right)}\left({y}^{\mathrm{4}} +{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +{y}+\mathrm{1}\right)} \\ $$$$=\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} −{y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)}{\left({y}^{\mathrm{4}} +{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +{y}+\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}\right)}{\mathrm{1}^{\mathrm{4}} +\mathrm{1}^{\mathrm{3}} +\mathrm{1}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$
Answered by Spillover last updated on 12/Jan/23
1+x=y x→0 y=1 lim_(y→1) ((y^6 −1)/(y^5 −1))=lim_(y→1) (((y^6 −1)/(y−1))/((y^5 −1)/(y−1))) from lim_(x→0) ((x^n −a^n )/(x−a))=na^(n−1) lim_(y→1) (((y^6 −1)/(y−1))/((y^5 −1)/(y−1)))=((6 ×1^(6−1) )/(5×1^(5−1) ))=(6/5)
$$\mathrm{1}+{x}={y}\:\:\:{x}\rightarrow\mathrm{0}\:\:\:{y}=\mathrm{1} \\ $$$$\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{y}^{\mathrm{6}} −\mathrm{1}}{{y}^{\mathrm{5}} −\mathrm{1}}=\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\frac{{y}^{\mathrm{6}} −\mathrm{1}}{{y}−\mathrm{1}}}{\frac{{y}^{\mathrm{5}} −\mathrm{1}}{{y}−\mathrm{1}}} \\ $$$${from}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}^{{n}} −{a}^{{n}} }{{x}−{a}}={na}^{{n}−\mathrm{1}} \\ $$$$\mathrm{li}\underset{{y}\rightarrow\mathrm{1}} {\mathrm{m}}\:\frac{\frac{{y}^{\mathrm{6}} −\mathrm{1}}{{y}−\mathrm{1}}}{\frac{{y}^{\mathrm{5}} −\mathrm{1}}{{y}−\mathrm{1}}}=\frac{\mathrm{6}\:×\mathrm{1}^{\mathrm{6}−\mathrm{1}} }{\mathrm{5}×\mathrm{1}^{\mathrm{5}−\mathrm{1}} }=\frac{\mathrm{6}}{\mathrm{5}} \\ $$

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