Question

$\boxed{solve \: this}$
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Answer 1

HELLO DEAR,    

$\bold{\lim_{x \to 0}\frac{tanx - sinx}{x^3}}$  

$\bold{\lim_{x \to 0} \frac{\frac{sinx}{cosx} - sinx}{x^3}}$

$\bold{\lim_{x \to 0} \frac{sinx(\frac{1}{cosx} - 1)}{x^3}}$

$\bold{\lim_{x \to 0} \frac{sinx(1 - cosx)}{x^3.cosx}}$

$\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{ 1 - cosx \over x^2 }}]$  

$\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{ 2sin^2{x\over 2} \over x^2 }}]$

$2\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{ sin{x\over 2} \over x}.{ sin{x\over 2} \over x}}]$

$2\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{1\over 2}{ sin{x\over 2} \over {x\over 2}}.{1\over 2}{ sin{x\over 2} \over {x\over 2}}}]$

$2[\lim_{x \to 0} {{1\over cos x}\times \lim_{x \to 0}{sin x \over x} \times \lim_{x \to 0}{1\over 2}{ sin{x\over 2} \over {x\over 2}}\times \lim_{x \to 0}{1\over 2}{sin{x\over 2} \over {x\over 2}}}]$  

$2[\lim_{x \to 0} {{1\over cos x}\times \lim_{x \to 0}{sin x \over x} \times \lim_{{x\over 2} \to 0}{1\over 2}{ sin{x\over 2} \over {x\over 2}}\times \lim_{{x\over2} \to 0}{1\over 2}{ sin{x\over 2} \over {x\over 2}}}] \\ \\ \bold{[as \: \: x \: \: tends \: \: to \: \: zero, \: \: x/2 \: \: also \: \: tends \: \: to \: \: zero]}$

$2\times 1 \times 1 \times {1 \over 2} \times {1 \over 2} = {1 \over 2}$    

I HOPE ITS HELP YOU DEAR,  

THANKS