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Evaluate the following

Evaluate the following

Question Image
Grade:12th pass

1 Answers

jagdish singh singh
173 Points
5 years ago
\hspace{-0.5 cm} $Let $\bf{\mathcal{A} = \lim_{x\rightarrow \infty}\frac{\sin x+\cos x}{x^2}\;,}$ Now Using $\bf{\sin x+\cos x = \sqrt{2}\sin \left(x+\frac{\pi}{4}\right)}$\\\\ \\And Using $\bf{-\sqrt{2} \leq \sqrt{2}\sin \left(x+\frac{\pi}{4}\right) \leq \sqrt{2}}$\\\\\\ So we get $\bf{-\lim_{x\rightarrow \infty}\frac{\sqrt{2}}{x^2}\leq \lim_{x\rightarrow \infty}\frac{\sqrt{2}\sin (x+\frac{\pi}{4})}{x^2}\leq \lim_{x\rightarrow \infty}\frac{\sqrt{2}}{x^2}}$\\\\\\
 
\hspace{-0.5 cm} $So we get $\bf{-0\leq \mathcal{A}\leq 0}$\\\\\\ So Using Squeeze Theorem, We get $\bf{\mathcal{A} =0}$

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