Find-lim-x-0-5x-tan-5x-x-3-

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 (\dots) = {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} + \dots)}{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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