A) 2
B) -2
C) 1
D) -1
Correct Answer: B
Solution :
[b] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ({{x}^{2}})}{In(cos(2{{x}^{2}}-x))}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ({{x}^{2}})}{\log \left( 1-2{{\sin }^{2}}\left( \frac{2{{x}^{2}}-x}{2} \right) \right)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ({{x}^{2}}){{x}^{2}}}{\frac{{{x}^{2}}\log \left( 1-2{{\sin }^{2}}\left( \frac{2{{x}^{2}}-x}{2} \right) \right)}{-2{{\sin }^{2}}\left( \frac{2{{x}^{2}}-x}{2} \right)}\left[ -2{{\sin }^{2}}\left( \frac{2{{x}^{2}}-x}{2} \right) \right]}=\]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\frac{2{{\sin }^{2}}\left( \frac{2{{x}^{2}}-x}{2} \right)}{{{\left( \frac{2{{x}^{2}}-x}{2} \right)}^{2}}}{{\left( \frac{2{{x}^{2}}-x}{2} \right)}^{2}}}\] =\[\underset{x\to 0}{\mathop{\lim }}\,-\frac{2{{x}^{2}}}{{{(2{{x}^{2}}-x)}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,-\frac{2}{{{(2x-1)}^{2}}}=-2\]
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