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Question Number 167105 by mnjuly1970 last updated on 06/Mar/22
If g(x)= { (( x^( 2) x≥1)),(( x^( 3) x< 1)) :} then lim_( h→ 0^( +) ) (( g (1+3h ) − g (1−5h ))/h) =?
$$ \\ $$$$\:\:\:\:\mathrm{I}{f}\:\:\:\:{g}\left({x}\right)=\:\begin{cases}{\:{x}^{\:\mathrm{2}} \:\:\:\:\:{x}\geqslant\mathrm{1}}\\{\:{x}^{\:\mathrm{3}} \:\:\:\:\:\:\:{x}<\:\mathrm{1}}\end{cases}\: \\ $$$$\:\:\:\:\:\:{then}\:\:\:\:\:{lim}_{\:{h}\rightarrow\:\mathrm{0}^{\:+} } \frac{\:{g}\:\left(\mathrm{1}+\mathrm{3}{h}\:\right)\:−\:{g}\:\left(\mathrm{1}−\mathrm{5}{h}\:\right)}{{h}}\:=? \\ $$
Answered by greogoury55 last updated on 06/Mar/22
lim_(x→0^+ ) ((g(1+3h)−g(1−5h))/h) = lim_(x→0^+ ) 3g ′(1+3h)+5 g ′(1−5h) = 8 g′(1)= 8×3×(1)^2 = 24
$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{g}\left(\mathrm{1}+\mathrm{3}{h}\right)−{g}\left(\mathrm{1}−\mathrm{5}{h}\right)}{{h}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\mathrm{3}{g}\:'\left(\mathrm{1}+\mathrm{3}{h}\right)+\mathrm{5}\:{g}\:'\left(\mathrm{1}−\mathrm{5}{h}\right) \\ $$$$\:=\:\mathrm{8}\:{g}'\left(\mathrm{1}\right)=\:\mathrm{8}×\mathrm{3}×\left(\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{24} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 06/Mar/22
please recheck
$$\:{please}\:{recheck} \\ $$
Answered by Mathspace last updated on 07/Mar/22
lim_(h→0^+ ) ((g(1+3h)−g(1−5h))/h) =lim_(h→o+) (((1+3h)^2 −(1−5h)^3 )/h) =lim_(h→0^+ ) 2.3(1+3h)−3.(−5) (1−5h)^2 =6+15 =21 (i have used hospital)
$${lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\:\frac{{g}\left(\mathrm{1}+\mathrm{3}{h}\right)−{g}\left(\mathrm{1}−\mathrm{5}{h}\right)}{{h}} \\ $$$$={lim}_{{h}\rightarrow{o}+} \:\:\:\frac{\left(\mathrm{1}+\mathrm{3}{h}\right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{5}{h}\right)^{\mathrm{3}} }{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\mathrm{2}.\mathrm{3}\left(\mathrm{1}+\mathrm{3}{h}\right)−\mathrm{3}.\left(−\mathrm{5}\right) \\ $$$$\left(\mathrm{1}−\mathrm{5}{h}\right)^{\mathrm{2}} \\ $$$$=\mathrm{6}+\mathrm{15}\:=\mathrm{21}\:\:\left({i}\:{have}\:{used}\:{hospital}\right) \\ $$
Answered by alephzero last updated on 08/Mar/22
g(x) = { (x^2 ,(x ≥ 1)),(x^3 ,(x < 1)) :} ⇒ lim_(x→+0) ((g(1+3x)−g(1−5x))/x) = = lim_(x→+0) (((3x+1)^3 −(5x−1)^3 )/x) (d/dx)(3x+1)^3 = (d/(d(3x+1)))(3x+1)^3 ×(d/dx)(3x+1) = = 9(3x+1)^2 (d/dx)(5x−1)^3 = (d/(d(5x−1)))(5x−1)^3 ×(d/dx)(5x−1) = = 15(5x−1)^2 ⇒ lim_(x→+0) (9(3x+1)^2 −15(5x−1)^3 ) = 9(1)^2 −15(−1)^3 = 9−(−15) = = 24
$${g}\left({x}\right)\:=\:\begin{cases}{{x}^{\mathrm{2}} }&{{x}\:\geqslant\:\mathrm{1}}\\{{x}^{\mathrm{3}} }&{{x}\:<\:\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow+\mathrm{0}} {\mathrm{lim}}\frac{{g}\left(\mathrm{1}+\mathrm{3}{x}\right)−{g}\left(\mathrm{1}−\mathrm{5}{x}\right)}{{x}}\:= \\ $$$$=\:\underset{{x}\rightarrow+\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{3}} }{{x}} \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{3}} \:=\:\frac{{d}}{{d}\left(\mathrm{3}{x}+\mathrm{1}\right)}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{3}} ×\frac{{d}}{{dx}}\left(\mathrm{3}{x}+\mathrm{1}\right)\:= \\ $$$$=\:\mathrm{9}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{3}} \:=\:\frac{{d}}{{d}\left(\mathrm{5}{x}−\mathrm{1}\right)}\left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{3}} ×\frac{{d}}{{dx}}\left(\mathrm{5}{x}−\mathrm{1}\right)\:= \\ $$$$=\:\mathrm{15}\left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow+\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{9}\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{15}\left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{3}} \right) \\ $$$$=\:\mathrm{9}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{15}\left(−\mathrm{1}\right)^{\mathrm{3}} \:=\:\mathrm{9}−\left(−\mathrm{15}\right)\:= \\ $$$$=\:\mathrm{24} \\ $$
Commented by cortano1 last updated on 08/Mar/22
(d/dx)(3x+1)^3 = 9(3x+1)^2
$$\:\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{3}} \:=\:\mathrm{9}\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Commented by alephzero last updated on 08/Mar/22
sorry for mistake! i′ve corrected now.
$$\mathrm{sorry}\:\mathrm{for}\:\mathrm{mistake}!\:\mathrm{i}'\mathrm{ve}\:\mathrm{corrected} \\ $$$$\mathrm{now}. \\ $$
Commented by mr W last updated on 08/Mar/22
it is (3x+1)^2 , not (3x+1)^3 ! you had many typos.
$${it}\:{is}\:\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} ,\:{not}\:\left(\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{3}} ! \\ $$$${you}\:{had}\:{many}\:{typos}. \\ $$
Answered by mr W last updated on 07/Mar/22
g(1+3h)=(1+3h)^2 =1+2×(3h)+o(h) g(1−5h)=(1−5h)^3 =1+3×(−5h)+o(h) lim_( h→ 0^( +) ) (( g (1+3h ) − g (1−5h ))/h) =lim_( h→ 0^( +) ) (( 1+6h+o(h) −1+15h+o(h))/h) =lim_( h→ 0^( +) ) (( 21h+o(h))/h) =21
$${g}\left(\mathrm{1}+\mathrm{3}{h}\right)=\left(\mathrm{1}+\mathrm{3}{h}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}×\left(\mathrm{3}{h}\right)+{o}\left({h}\right) \\ $$$${g}\left(\mathrm{1}−\mathrm{5}{h}\right)=\left(\mathrm{1}−\mathrm{5}{h}\right)^{\mathrm{3}} =\mathrm{1}+\mathrm{3}×\left(−\mathrm{5}{h}\right)+{o}\left({h}\right) \\ $$$${lim}_{\:{h}\rightarrow\:\mathrm{0}^{\:+} } \frac{\:{g}\:\left(\mathrm{1}+\mathrm{3}{h}\:\right)\:−\:{g}\:\left(\mathrm{1}−\mathrm{5}{h}\:\right)}{{h}} \\ $$$$={lim}_{\:{h}\rightarrow\:\mathrm{0}^{\:+} } \frac{\:\mathrm{1}+\mathrm{6}{h}+{o}\left({h}\right)\:−\mathrm{1}+\mathrm{15}{h}+{o}\left({h}\right)}{{h}} \\ $$$$={lim}_{\:{h}\rightarrow\:\mathrm{0}^{\:+} } \frac{\:\mathrm{21}{h}+{o}\left({h}\right)}{{h}} \\ $$$$=\mathrm{21} \\ $$

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