Question

the value of limited x to 0 sin7x+sin5x÷tan5x-tan2x is​

Answer 1

$\underline{\textsf{To find:}}$

$\textsf{The value of}$

$\mathsf{\displaystyle\lim_{x\to\;0}\;\dfrac{sin7x+sin5x}{tan5x-tan2x}}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\displaystyle\lim_{x\to\;0}\;\dfrac{sin7x+sin5x}{tan5x-tan2x}}$

$\mathsf{when\;applying\;limits\;we\;get\;\dfrac{0}{0}\;form}$

$\mathsf{By\;L\;Hopital,s\;rule}$

$\mathsf{=\displaystyle\lim_{x\to\;0}\;\dfrac{7\;cos7x+5\;cos5x}{5\;sec^25x-2\;sec^22x}}$

$\mathsf{=\dfrac{7\;cos0+5\;cos0}{5\;sec^20-2\;sec^20}}$

$\mathsf{=\dfrac{7(1)+5(1)}{5(1)-2(1)}}$

$\mathsf{=\dfrac{7+5}{5-2}}$

$\mathsf{=\dfrac{12}{3}}$

$\mathsf{=4}$

$\therefore\boxed{\mathsf{\displaystyle\lim_{x\to\;0}\;\dfrac{sin7x+sin5x}{tan5x-tan2x}=4}}$

$\underline{\textsf{Find more:}}$

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Answer 2

SOLUTION :-

TO DETERMINE :-

$\displaystyle \sf{ \lim_{x \to 0} \: \: \frac{ \sin 7x + \sin 5x}{ \tan 5x - \tan 2x}}$

EVALUATION :-

$\displaystyle \sf{ \lim_{x \to 0} \: \: \frac{ \sin 7x + \sin 5x}{ \tan 5x - \tan 2x}}$

$= \displaystyle \sf{ \lim_{x \to 0} \: \: \frac{2 \sin \big( \frac{7x + 5x}{2} \big) \cos \big( \frac{7x - 5x}{2} \big) }{ \frac{ \sin 5x}{ \cos 5x} - \frac{ \sin 2x}{ \cos 2x} }}$

$= \displaystyle \sf{ \lim_{x \to 0} \: \: \frac{2 \sin 6x \cos x }{ \frac{ \sin 5x\cos 2x - \cos 5x \sin 2x}{ \cos 5x\cos 2x} }}$

$= \displaystyle \sf{ \lim_{x \to 0} \: \: \frac{2 . 2 \sin 3x . \cos 3x \cos x }{ \frac{ \sin (5x - 2x) }{ \cos 5x\cos 2x} }}$

$= \displaystyle \sf{ \lim_{x \to 0} \: \: \frac{4 \sin 3x \cos 3x \cos x \cos 5x\cos 2x }{ { \sin 3x } }}$

$= \displaystyle \sf{ \lim_{x \to 0} \: \: {4 \cos 3x \cos x \cos 5x \cos 2x}}$

$= \displaystyle \sf{ \: \: {4 \cos 0.\cos 0. \cos 0 .\cos 0}}$

$= \displaystyle \sf{ 4 \times 1 \times 1 \times 1 \times 1}$

$= \displaystyle \sf{4}$

FINAL ANSWER :-

$\boxed{ \: \boxed{ \: \: \displaystyle \sf{ \lim_{x \to 0} \: \: \frac{ \sin 7x + \sin 5x}{ \tan 5x - \tan 2x} = 4 \: \: }} \:}$

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