lim _{x rightarrow infty} frac{3 x+4 cos ^2 x | vedclass

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(C) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}}$,we divide the numerator and the denominator by

Vedclass · Accepted Answer

$3$

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(C) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}}$,we divide the numerator and the denominator by $x$ (since $x \rightarrow \infty$,$x > 0$,so $\sqrt{x^2} = x$): $\lim _{x \rightarrow \infty} \frac{3 + \frac{4 \cos ^2 x}{x}}{\sqrt{\frac{x^2}{x^2} - \frac{5 \sin ^2 x}{x^2}}} = \lim _{x \rightarrow \infty} \frac{3 + \frac{4 \cos ^2 x}{x}}{\sqrt{1 - \frac{5 \sin ^2 x}{x^2}}}$ As $x \rightarrow \infty$,the terms $\frac{4 \cos ^2 x}{x} \rightarrow 0$ (because $\cos ^2 x$ is bounded between $0$ and $1$) and $\frac{5 \sin ^2 x}{x^2} \rightarrow 0$ (because $\sin ^2 x$ is bounded between $0$ and $1$). Thus,the limit becomes $\frac{3 + 0}{\sqrt{1 - 0}} = \frac{3}{1} = 3$.

$\lim _{x \rightarrow \infty} \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}} = $

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