If g(x)= { (( x^( 2) x≥1)),(( x^( 3) x< 1)) :} then lim_( h→ 0^( +) ) (( g (1+3h ) − g (1−5h ))/h) =?


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Question Number 167105 by mnjuly1970 last updated on 06/Mar/22
$If g(x)= { (( x^( 2) x≥1)),(( x^( 3) x< 1)) :} then lim_( h→ 0^( +) ) (( g (1+3h ) − g (1−5h ))/h) =?$
$I f g (x) = {\begin{cases} x^{2} x ⩾ 1 \\ x^{3} x < 1 \end{cases}$ $t h e n {l i m}_{h \to 0^{+}} \frac{g (1 + 3 h) - g (1 - 5 h)}{h} = ?$
	Answered by greogoury55 last updated on 06/Mar/22

	$\lim_{x \to 0^{+}} \frac{g (1 + 3 h) - g (1 - 5 h)}{h}$ $= \lim_{x \to 0^{+}} 3 g^{'} (1 + 3 h) + 5 g^{'} (1 - 5 h)$ $= 8 g^{'} (1) = 8 \times 3 \times {(1)}^{2} = 24$
	Commented bymnjuly1970 last updated on 06/Mar/22

	$p l e a s e r e c h e c k$
	Answered by Mathspace last updated on 07/Mar/22

	${l i m}_{h \to 0^{+}} \frac{g (1 + 3 h) - g (1 - 5 h)}{h}$ $= {l i m}_{h \to o +} \frac{{(1 + 3 h)}^{2} - {(1 - 5 h)}^{3}}{h}$ $= {l i m}_{h \to 0^{+}} 2.3 (1 + 3 h) - 3 . (- 5)$ ${(1 - 5 h)}^{2}$ $= 6 + 15 = 21 (i h a v e u s e d h o s p i t a l)$
	Answered by alephzero last updated on 08/Mar/22
	$g(x) = { (x^2 ,(x ≥ 1)),(x^3 ,(x < 1)) :} ⇒ lim_(x→+0) ((g(1+3x)−g(1−5x))/x) = = lim_(x→+0) (((3x+1)^3 −(5x−1)^3 )/x) (d/dx)(3x+1)^3 = (d/(d(3x+1)))(3x+1)^3 ×(d/dx)(3x+1) = = 9(3x+1)^2 (d/dx)(5x−1)^3 = (d/(d(5x−1)))(5x−1)^3 ×(d/dx)(5x−1) = = 15(5x−1)^2 ⇒ lim_(x→+0) (9(3x+1)^2 −15(5x−1)^3 ) = 9(1)^2 −15(−1)^3 = 9−(−15) = = 24$
	$g (x) = {\begin{cases} x^{2} & x ⩾ 1 \\ x^{3} & x < 1 \end{cases}$ $\Rightarrow \lim_{x \to + 0} \frac{g (1 + 3 x) - g (1 - 5 x)}{x} =$ $= \lim_{x \to + 0} \frac{{(3 x + 1)}^{3} - {(5 x - 1)}^{3}}{x}$ $\frac{d}{d x} {(3 x + 1)}^{3} = \frac{d}{d (3 x + 1)} {(3 x + 1)}^{3} \times \frac{d}{d x} (3 x + 1) =$ $= 9 {(3 x + 1)}^{2}$ $\frac{d}{d x} {(5 x - 1)}^{3} = \frac{d}{d (5 x - 1)} {(5 x - 1)}^{3} \times \frac{d}{d x} (5 x - 1) =$ $= 15 {(5 x - 1)}^{2}$ $\Rightarrow \lim_{x \to + 0} (9 {(3 x + 1)}^{2} - 15 {(5 x - 1)}^{3})$ $= 9 {(1)}^{2} - 15 {(- 1)}^{3} = 9 - (- 15) =$ $= 24$
	Commented bycortano1 last updated on 08/Mar/22

	$\frac{d}{dx} {(3 x + 1)}^{3} = 9 {(3 x + 1)}^{2}$
	Commented byalephzero last updated on 08/Mar/22

	$sorry for mistake! i^{'} ve corrected$ $now .$
	Commented bymr W last updated on 08/Mar/22

	$i t i s {(3 x + 1)}^{2}, n o t {(3 x + 1)}^{3}!$ $y o u h a d m a n y t y p o s .$
	Answered by mr W last updated on 07/Mar/22

	$g (1 + 3 h) = {(1 + 3 h)}^{2} = 1 + 2 \times (3 h) + o (h)$ $g (1 - 5 h) = {(1 - 5 h)}^{3} = 1 + 3 \times (- 5 h) + o (h)$ ${l i m}_{h \to 0^{+}} \frac{g (1 + 3 h) - g (1 - 5 h)}{h}$ $= {l i m}_{h \to 0^{+}} \frac{1 + 6 h + o (h) - 1 + 15 h + o (h)}{h}$ $= {l i m}_{h \to 0^{+}} \frac{21 h + o (h)}{h}$ $= 21$
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