lim _{x rightarrow infty}left(frac{6 x^2-cos | vedclass

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(D) We evaluate the limit as $x \rightarrow \infty$: $\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\righ

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(D) We evaluate the limit as $x \rightarrow \infty$: $\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right)$ Divide the numerator and denominator of the first term by $x^2$: $\lim _{x \rightarrow \infty} \frac{6 - \frac{\cos 3x}{x^2}}{1 + \frac{5}{x^2}} = \frac{6 - 0}{1 + 0} = 6$ For the second term,since $x \rightarrow \infty$,$x > 0$,so $\sqrt{x^6} = x^3$: $\lim _{x \rightarrow \infty} \frac{5x^3 + 3}{\sqrt{x^6+2}} = \lim _{x \rightarrow \infty} \frac{5 + \frac{3}{x^3}}{\sqrt{1 + \frac{2}{x^6}}} = \frac{5 + 0}{\sqrt{1 + 0}} = 5$ Thus,the limit is $6 - 5 = 1$.

$\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right) = $

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