Menu Close

Question-173765

Tinku Tara 0 Comments
Pin




Question Number 173765 by saly last updated on 17/Jul/22
Answered by cortano1 last updated on 18/Jul/22
L = lim_(x→1) ((3/(1−(√x))) − (2/(1−(x)^(1/3) )) )=? [ let x = t^6 ∧ t→1 ] L= lim_(t→1) ((3/(1−t^3 )) − (2/(1−t^2 )) ) = lim_(t→1) ((3/((1−t)(1+t+t^2 ))) −(2/((1−t)(1+t)))) = lim_(t→1) (((3t+3−2t^2 −2t−2)/((1−t)(1+t)(1+t+t^2 )))) = (1/6)×lim_(t→1) (((−(t−1)(2t+1))/(−(t−1)))) = (1/6)×3 = (1/2)
$$\:{L}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{3}}{\mathrm{1}−\sqrt{{x}}}\:−\:\frac{\mathrm{2}}{\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}}\:\right)=? \\ $$$$\:\left[\:{let}\:{x}\:=\:{t}^{\mathrm{6}} \:\wedge\:{t}\rightarrow\mathrm{1}\:\right] \\ $$$$\:{L}=\:\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{3}}{\mathrm{1}−{t}^{\mathrm{3}} }\:−\:\frac{\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\:\:=\:\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{3}}{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}\:−\frac{\mathrm{2}}{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}\right) \\ $$$$\:\:\:\:\:=\:\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\mathrm{3}{t}+\mathrm{3}−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{2}}{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}\right) \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{6}}×\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{−\left({t}−\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)}{−\left({t}−\mathrm{1}\right)}\right) \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{3}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by saly last updated on 18/Jul/22
Thank you
$${Thank}\:{you} \\ $$
Answered by blackmamba last updated on 18/Jul/22
set x−1 =r ∧ r→0 = lim_(r→0) ((3/(1−(√(1+r)))) −(2/(1−((1+r))^(1/3) ))) = lim_(r→0) (((3(1−((1+r))^(1/3) )−2(1−(√(1+r))))/((1−(√(1+r)))(1−((1+r))^(1/3) )))) =lim_(r→0) (((3(1−(1+(r/3)−(1/9)r^2 ))−2(1−(1+(1/2)r−(1/8)r^2 )))/((1−(1+(1/2)r))(1−(1+(1/3)r)))) = lim_(r→0) (((3(−(r/3)+(r^2 /9))−2(−(r/2)+(r^2 /8)))/((−(1/2)r)(−(1/3)r)))) = lim_(r→0) ((((r^2 /3)−(r^2 /4))/((1/6)r^2 )))= 6×(((4−3)/(12)))= (1/2)
$$\:{set}\:{x}−\mathrm{1}\:={r}\:\wedge\:{r}\rightarrow\mathrm{0} \\ $$$$\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}}{\mathrm{1}−\sqrt{\mathrm{1}+{r}}}\:−\frac{\mathrm{2}}{\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{1}+{r}}}\right) \\ $$$$\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{1}+{r}}\right)−\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{1}+{r}}\right)}{\left(\mathrm{1}−\sqrt{\mathrm{1}+{r}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{1}+{r}}\right)}\right) \\ $$$$\:\:=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}\left(\mathrm{1}−\left(\mathrm{1}+\frac{{r}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{9}}{r}^{\mathrm{2}} \right)\right)−\mathrm{2}\left(\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{r}−\frac{\mathrm{1}}{\mathrm{8}}{r}^{\mathrm{2}} \right)\right)}{\left(\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{r}\right)\right)\left(\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}{r}\right)\right)}\right. \\ $$$$\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}\left(−\frac{{r}}{\mathrm{3}}+\frac{{r}^{\mathrm{2}} }{\mathrm{9}}\right)−\mathrm{2}\left(−\frac{{r}}{\mathrm{2}}+\frac{{r}^{\mathrm{2}} }{\mathrm{8}}\right)}{\left(−\frac{\mathrm{1}}{\mathrm{2}}{r}\right)\left(−\frac{\mathrm{1}}{\mathrm{3}}{r}\right)}\right) \\ $$$$\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\frac{{r}^{\mathrm{2}} }{\mathrm{3}}−\frac{{r}^{\mathrm{2}} }{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{6}}{r}^{\mathrm{2}} }\right)=\:\mathrm{6}×\left(\frac{\mathrm{4}−\mathrm{3}}{\mathrm{12}}\right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by CElcedricjunior last updated on 18/Jul/22
−(1/2) .....le celebre cedric junior......
$$−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$…..{le}\:{celebre}\:{cedric}\:{junior}…… \\ $$
Answered by Mathspace last updated on 18/Jul/22
f(x)=(3/(1−(√x)))−(2/(1−^3 (√x))) we do the cha7gement x=1−t (so t→0) ⇒ f(x)=f(1−t)=(3/(1−(√(1−t))))−(2/(1−^3 (√x))) (1+u)^α ∼1+αu +((α(α−1))/2)u^2 (1−u)^α ∼1−αu+((α(α−1))/2)u^2 (1−t)^(1/2) ∼1−(t/2)−(1/8)t^2 (1−t)^(1/3) ∼1−(t/3)−(2/(18))t^2 ⇒f(1−t)∼(3/((t/2)+(t^2 /8)))−(2/((t/3)+(t^2 /9))) =((24)/(4t+t^2 ))−((18)/(3t+t^2 )) =((24(3t+t^2 )−18(4t+t^2 ))/((4t+t^2 )(3t+t^2 ))) =((72t+24t^2 −72t−18t^2 )/(t^2 (4+t)(3+t))) =(6/((4+t)(3+t)))→(6/(12))=(1/2)(t→0)
$${f}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{1}−\sqrt{{x}}}−\frac{\mathrm{2}}{\mathrm{1}−^{\mathrm{3}} \sqrt{{x}}} \\ $$$${we}\:{do}\:{the}\:{cha}\mathrm{7}{gement}\:{x}=\mathrm{1}−{t} \\ $$$$\left({so}\:{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)={f}\left(\mathrm{1}−{t}\right)=\frac{\mathrm{3}}{\mathrm{1}−\sqrt{\mathrm{1}−{t}}}−\frac{\mathrm{2}}{\mathrm{1}−^{\mathrm{3}} \sqrt{{x}}} \\ $$$$\left(\mathrm{1}+{u}\right)^{\alpha} \sim\mathrm{1}+\alpha{u}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}}{u}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{u}\right)^{\alpha} \sim\mathrm{1}−\alpha{u}+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}}{u}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \sim\mathrm{1}−\frac{{t}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}{t}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \sim\mathrm{1}−\frac{{t}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{18}}{t}^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{1}−{t}\right)\sim\frac{\mathrm{3}}{\frac{{t}}{\mathrm{2}}+\frac{{t}^{\mathrm{2}} }{\mathrm{8}}}−\frac{\mathrm{2}}{\frac{{t}}{\mathrm{3}}+\frac{{t}^{\mathrm{2}} }{\mathrm{9}}} \\ $$$$=\frac{\mathrm{24}}{\mathrm{4}{t}+{t}^{\mathrm{2}} }−\frac{\mathrm{18}}{\mathrm{3}{t}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{24}\left(\mathrm{3}{t}+{t}^{\mathrm{2}} \right)−\mathrm{18}\left(\mathrm{4}{t}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{4}{t}+{t}^{\mathrm{2}} \right)\left(\mathrm{3}{t}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{72}{t}+\mathrm{24}{t}^{\mathrm{2}} −\mathrm{72}{t}−\mathrm{18}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left(\mathrm{4}+{t}\right)\left(\mathrm{3}+{t}\right)} \\ $$$$=\frac{\mathrm{6}}{\left(\mathrm{4}+{t}\right)\left(\mathrm{3}+{t}\right)}\rightarrow\frac{\mathrm{6}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}}\left({t}\rightarrow\mathrm{0}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *