Lmlit x→0 (tan3X-sin3X)/X^3=?

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Answer 1 · 2018-06-28T19:48:03

$lim_(xto0)(tan3x-sin3x)/x^3=27/2$

Let ,

$L=lim_(xto0)(tan3x-sin3x)/x^3$

$L=lim_(xto0)((sin3x)/(cos3x)-sin3x)/x^3$

$L=lim_(xto0)(sin3x(1/(cos3x)-1))/x^3$

$L=lim_(xto0)(sin3x(1-cos3x))/(x^3cos3x)to[because1-costheta=2sin^2(theta/2)]$

$L=lim_(xto0)(sin3x)/x*lim_(xto0)(2sin^2((3x)/2))/x^2*lim_(xto0)1/(cos3x)$

$L=(3lim_(3xto0)(sin3x)/(3x)) (2lim_((3x)/2to0)[sin((3x)/2)/((3x)/2)]^2 *9/4)(1/cos0)$

$L$ = $(3*1)(2(1)^2*9/4)(1)to[because lim_(theta to0) sintheta/theta =1, theta=3x, (3x)/2]$

$L=3xx2xx9/4$

$L=27/2$

Answer 2 · 2018-06-28T19:53:13

$27/2$

Using Taylor's series expansions of $\tan 3x$ & $\sin 3x$
as follows

$\lim_{x\to 0}\frac{\tan 3x-\sin 3x}{x^3}$

$=\lim_{x\to 0}\frac{(3x+\frac{(3x)^3}{3}+\frac{2(2x)^5}{15}+\ldots)-(3x-\frac{(3x)^3}{3!}+\frac{(2x)^5}{5!}+\ldots)}{x^3}$

$=\lim_{x\to 0}\frac{\frac{(3x)^3}{2}+\frac{(2x)^5}{8}+\ldots}{x^3}$

$=\lim_{x\to 0}\frac{x^3(\frac{27}{2}+4x^2+\ldots)}{x^3}$

$=\lim_{x\to 0}(\frac{27}{2}+4x^2+\ldots)$

$=\frac{27}{2}+0$

$=27/2$