Question-160142

Question Number 160142 by cortano last updated on 25/Nov/21

Answered by puissant last updated on 25/Nov/21

lim_(x→1) ((x^(1111) +2x^(2222) −3x^(3333) )/(x^(4444) −1)) = lim_(x→1) (((d/dx){x^(1111) +2x^(2222) −3x^(3333) })/((d/dx){x^(4444) −1})) = lim_(x→1) ((1111x^(1110) +4444x^(2221) −9999x^(3332) )/(4444x^(4443) )) = ((1111+4444−9999)/(4444)) = −((4444)/(4444)) = −1.. ............Le puissant............

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}^{\mathrm{1111}} +\mathrm{2}{x}^{\mathrm{2222}} −\mathrm{3}{x}^{\mathrm{3333}} }{{x}^{\mathrm{4444}} −\mathrm{1}}\: \\ $$$$=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left\{{x}^{\mathrm{1111}} +\mathrm{2}{x}^{\mathrm{2222}} −\mathrm{3}{x}^{\mathrm{3333}} \right\}}{\frac{{d}}{{dx}}\left\{{x}^{\mathrm{4444}} −\mathrm{1}\right\}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1111}{x}^{\mathrm{1110}} +\mathrm{4444}{x}^{\mathrm{2221}} −\mathrm{9999}{x}^{\mathrm{3332}} }{\mathrm{4444}{x}^{\mathrm{4443}} } \\ $$$$=\:\frac{\mathrm{1111}+\mathrm{4444}−\mathrm{9999}}{\mathrm{4444}}\:=\:−\frac{\mathrm{4444}}{\mathrm{4444}}\:=\:−\mathrm{1}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:…………\mathscr{L}{e}\:{puissant}………… \\ $$

Commented by cortano last updated on 25/Nov/21

$${yes} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Nov/21

≻_( ⋏) ^( ⋎) Without Using Calculus_⋏ ^⋎ ≺ lim((x^(1111) +2x^(2222) −3x^(3333) )/(x^(4444) −1)) lim_(x→1) ((x^(1111) −x^(2222) +3x^(2222) −3x^(3333) )/(x^(4444) −1)) lim_(x→1) ((x^(1111) (1−x^(1111) )+3x^(2222) (1−x^(1111) ))/((x^(2222) −1)(x^(2222) +1))) lim_(x→1) ((−(x^(1111) −1)(x^(1111) +3x^(2222) ))/((x^(1111) −1)(x^(1111) +1)(x^(2222) +1))) lim_(x→1) ((−(x^(1111) +3x^(2222) ))/((x^(1111) +1)(x^(2222) +1)))=((−(1+3))/((1+1)(1+1))) =((−4)/4)=−1

$$\succ_{\:\curlywedge} ^{\:\curlyvee} \mathrm{Without}\:\mathrm{Using}\:\mathrm{Calculus}_{\curlywedge} ^{\curlyvee} \prec \\ $$$$\mathrm{lim}\frac{{x}^{\mathrm{1111}} +\mathrm{2}{x}^{\mathrm{2222}} −\mathrm{3}{x}^{\mathrm{3333}} }{{x}^{\mathrm{4444}} −\mathrm{1}} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}^{\mathrm{1111}} −{x}^{\mathrm{2222}} +\mathrm{3}{x}^{\mathrm{2222}} −\mathrm{3}{x}^{\mathrm{3333}} }{{x}^{\mathrm{4444}} −\mathrm{1}} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}^{\mathrm{1111}} \left(\mathrm{1}−{x}^{\mathrm{1111}} \right)+\mathrm{3}{x}^{\mathrm{2222}} \left(\mathrm{1}−{x}^{\mathrm{1111}} \right)}{\left({x}^{\mathrm{2222}} −\mathrm{1}\right)\left({x}^{\mathrm{2222}} +\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{−\cancel{\left({x}^{\mathrm{1111}} −\mathrm{1}\right)}\left({x}^{\mathrm{1111}} +\mathrm{3}{x}^{\mathrm{2222}} \right)}{\cancel{\left({x}^{\mathrm{1111}} −\mathrm{1}\right)}\left({x}^{\mathrm{1111}} +\mathrm{1}\right)\left({x}^{\mathrm{2222}} +\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{−\left({x}^{\mathrm{1111}} +\mathrm{3}{x}^{\mathrm{2222}} \right)}{\left({x}^{\mathrm{1111}} +\mathrm{1}\right)\left({x}^{\mathrm{2222}} +\mathrm{1}\right)}=\frac{−\left(\mathrm{1}+\mathrm{3}\right)}{\left(\mathrm{1}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)} \\ $$$$=\frac{−\mathrm{4}}{\mathrm{4}}=−\mathrm{1} \\ $$

Commented by puissant last updated on 25/Nov/21

Sir Rasheed.Sindhi (1−x^(1111) )=−(x^(1111) −1)

$${Sir}\:{Rasheed}.{Sindhi}\:\left(\mathrm{1}−{x}^{\mathrm{1111}} \right)=−\left({x}^{\mathrm{1111}} −\mathrm{1}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 25/Nov/21

Th∀nks to correct me sir! I′m going to edit my answer.

$$\mathbb{T}\mathrm{h}\forall\mathrm{n}\Bbbk\mathrm{s}\:\mathrm{to}\:\mathrm{correct}\:\mathrm{me}\:\mathrm{sir}!\:\mathrm{I}'\mathrm{m}\:\mathrm{going} \\ $$$$\mathrm{to}\:\mathrm{edit}\:\mathrm{my}\:\mathrm{answer}. \\ $$

Commented by cortano last updated on 25/Nov/21

$${okay} \\ $$

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