Limx - 0sinx2lncos2x2-x is equal to

Lim_x → 0sin(x^2)ln(cos(2x^2 x)) nbsp; lim_x → 0sin(x^2)log(1 2sin^2(2x^2 x2)) nbsp; lim_x → 0sin(x^2)x^2x^2 log(1 2sin^2(2x^2 x2)) 2 sin^2 ( 2x^2 x2)[ 2 ...

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Byju's Answer

Standard XIII

Mathematics

Standard Limits to Remove Indeterminate Form

limx → 0sinx2...

Question

$lim x \to 0 \frac{sin (x^{2})}{ln (cos (2 x^{2} - x))}$ is equal to

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Solution

The correct option is B $- 2$
$lim x \to 0 \frac{sin (x^{2})}{ln (cos (2 x^{2} - x))}$
$lim x \to 0 \frac{sin (x^{2})}{log (1 - 2 {sin}^{2} (\frac{2 x^{2} - x}{2}))}$
$lim x \to 0 \frac{sin (x^{2}) x^{2}}{\frac{x^{2} log (1 - 2 {sin}^{2} (\frac{2 x^{2} - x}{2}))}{- 2 {sin}^{2} (\frac{2 x^{2} - x}{2})} [- 2 {sin}^{2} (\frac{2 x^{2} - x}{2})]}$
$= lim x \to 0 - \frac{x^{2}}{\frac{2 {sin}^{2} (\frac{2 x^{2} - x}{2})}{{(\frac{2 x^{2} - x}{2})}^{2}}} \frac{1}{{(\frac{2 x^{2} - x}{2})}^{2}}$
$= lim x \to 0 - \frac{2 x^{2}}{(2 x^{2} - x)^{2}} = lim x \to 0 - \frac{2}{(2 x - 1)^{2}} = - 2$

Suggest Corrections

limx → 0sin(x2)ln(cos(2x2−x)) is equal to

$lim x \to 0 \frac{sin (x^{2})}{ln (cos (2 x^{2} - x))}$ is equal to