BeMath-Without-L-Hopital-and-series-find-lim-x-0-tan-x-x-x-3-

Question Number 107536 by bemath last updated on 11/Aug/20

♠ B e M a t h ★

W i t h o u t L^{'} H o p i t a l a n d s e r i e s

f i n d \lim_{x \to 0} \frac{\tan x - x}{x^{3}}

Commented by bobhans last updated on 12/Aug/20

just comparing with L^{'} Hopital

\lim_{x \to 0} \frac{\sec^{2} x - 1}{{3 x}^{2}} = \lim_{x \to 0} \frac{1 - \cos^{2} x}{{3 x}^{2} \cos^{2} x}

= \lim_{x \to 0} \frac{\sin^{2} x}{{3 x}^{2}} \times \lim_{x \to 0} \frac{1}{\cos^{2} x} = \frac{1}{3} \times 1 = \frac{1}{3}

Answered by $@y@m last updated on 12/Aug/20

L e t x = 2 y

L = \lim_{x \to 0} \frac{\tan x - x}{x^{3}}

L = \lim_{y \to 0} \frac{\tan 2 y - 2 y}{8 y^{3}}

= \lim_{y \to 0} \frac{\frac{2 \tan y}{1 - \tan^{2} y} - 2 y}{8 y^{3}}

= \lim_{y \to 0} {\frac{\tan y - y (1 - \tan^{2} y)}{4 y^{3} (1 - \tan^{2} y)}}

= \lim_{y \to 0} {\frac{\tan y - y + y \tan^{2} y}{4 y^{3} (1 - \tan^{2} y)}}

= \lim_{y \to 0} {\frac{\tan y - y}{4 y^{3} (1 - \tan^{2} y)}} + \lim_{y \to 0} {\frac{y \tan^{2} y}{4 y^{3} (1 - \tan^{2} y)}}

L = \frac{L}{4} + \lim_{y \to 0} {\frac{\tan^{2} y}{4 y^{2} (1 - \tan^{2} y)}}

L = \frac{L}{4} + \frac{1}{4} {\lim_{y \to 0} (\frac{\tan^{2} y}{y^{2}}) \times \lim_{y \to 0} \frac{1}{1 - \tan^{2} y}}

\frac{3 L}{4} = \frac{1}{4} \times 1

L = \frac{1}{3}

Commented by john santu last updated on 11/Aug/20

c o o l l \dots s i r

Commented by bemath last updated on 11/Aug/20

t h a n k y o u s i r

Commented by malwaan last updated on 11/Aug/20

e x c u s e m e s i r

a t t h e t h i r d l i n e f r o m d o w n

L = \frac{L}{4} + \frac{1}{4} \lim_{y \to 0} {(\frac{{t a n}^{2} y}{y^{2}}) \times \lim_{y \to 0} \frac{1}{1 - \tan^{2} y}}

\frac{3 L}{4} = \frac{1}{4} \times 1

L = \frac{1}{3}

t h a n k s

Commented by $@y@m last updated on 12/Aug/20

Answered by mathmax by abdo last updated on 11/Aug/20

try to not imposing conditions for solving a question \dots!

Commented by $@y@m last updated on 11/Aug/20

I d i s a g r e e .

I t m a k e s q u e s t i o n m o r e c h a l l a n g i n g .

Commented by mathmax by abdo last updated on 11/Aug/20

there is no challenge in this forum there is no prize \dots

Commented by bemath last updated on 12/Aug/20

s i r A b d o, w h y w i t h m y q u e s t i o n ? i s t h e r e s o m e t h i n g w r o n g ? o r a n y o n e o b j e c t t o t h a t q u e s t i o n ?

Commented by $@y@m last updated on 12/Aug/20

No one is here for prizes.