Find lim_(x→0) ((5x−tan (5x))/x^3 )


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Question Number 28341 by Cheyboy last updated on 24/Jan/18

$F i n d \lim_{x \to 0} \frac{5 x - \tan (5 x)}{x^{3}}$
Commented by Cheyboy last updated on 24/Jan/18

$\lim_{x \to 0} \frac{5 - \sec^{2} (5 x) . 5}{3 x^{2}}$ $\lim_{x \to 0} \frac{0 - 50 (\sec (5 x)) . \sec (5 x) . \sec (5 x) \tan (5 x)}{6 x}$ $\lim_{x \to 0} \frac{- {50 \sec}^{3} (5 x) \tan (5 x)}{6 x}$ $\lim_{x \to 0} \frac{- 50 (\sec^{5} (5 x) . 5 + {5 \tan}^{2} (5 x) \sec^{} (5 x))}{6}$ $\lim_{x \to 0} \frac{- 250 (1)}{6} = \frac{- 125}{3}$ $t h a n k s i r i w a s w i t h p r e s s u r e t h a t$ $y i c o u l d n t s o l v e i t i n c l a s s$
Commented by abdo imad last updated on 25/Jan/18

$l e t u s e t h e c h 5 x = t s o x \to 0 \Leftrightarrow t \to 0$ $l i m (. . .) = {l i m}_{t \to 0} \frac{t - t a n t}{\frac{t^{3}}{125}} = 125 {l i m}_{t \to 0^{}} \frac{t - t a n t}{t^{3}}$ $b u t {l i m}_{t \to 0} \frac{t - t a n t}{t^{3}} = {l i m}_{t \to 0} \frac{1 - (1 + {t a n}^{2} t)}{3 t^{2}}$ $= {l i m}_{t \to 0} \frac{- 2 t a n t (1 + {t a n}^{2} t)}{6 t} = \frac{- 1}{3} {l i m}_{t \to 0} \frac{t a n t}{t} = \frac{- 1}{3} \times 1 = \frac{- 1}{3}$ $\Rightarrow {l i m}_{x \to 0} \frac{5 x - t a n (5 x)}{x^{3}} = \frac{- 125}{3} .$
Commented by abdo imad last updated on 25/Jan/18

$i h a v e u s e d h o s p i t a l t h e o r e m .$
Commented by Cheyboy last updated on 25/Jan/18

$T h a n k s i r G o d b l e s s y o u$ $=$
	Answered by ajfour last updated on 24/Jan/18

	$= \lim_{x \to 0} \frac{5 x - (5 x + \frac{{(5 x)}^{3}}{3} + \frac{2 {(5 x)}^{5}}{15} + . . .)}{x^{3}}$ $= \lim_{x \to 0} \frac{- \frac{125}{3} + \frac{2 \times 125}{15} {(5 x)}^{2}}{1}$ $= - \frac{125}{3} .$
	Commented by Cheyboy last updated on 24/Jan/18

	$o o h s i r t h e n i m i s s t h e a n s$ $i t w a s p a r t o f o u r t e s t 2 d a y$ $b u t i g o t \frac{- 25}{2} o k t h a n k f o r t h e a n s w e r$
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