Question

friends, please answer this question ​

Answer 1

Answer:

answer for the given problem is given

Answer 2

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QUESTION:

$\displaystyle \lim_{x \to \infty} \frac{2x + 7 \sin x}{4x + 2 \cos x}$

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ANSWER:

$\displaystyle \lim_{x \to \infty} \frac{2x + 7 \sin x}{4x + 2 \cos x} = \frac{1}{2}$

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GIVEN:

$\displaystyle \lim_{x \to \infty} \frac{2x + 7 \sin x}{4x + 2 \cos x}$

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TO FIND:

$The \ value \ of \ \displaystyle \lim_{x \to \infty} \frac{2x + 7 \sin x}{4x + 2 \cos x}$

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EXPLANATION:

$\displaystyle \lim_{x \to \infty} \frac{2x + 7 \sin x}{4x + 2 \cos x} \\ \\ \\ \displaystyle \frac{2 (\infty) + 7 \sin (\infty)}{4(\infty) + 2 \cos (\infty)} \\ \\ \\ \frac{\infty}{\infty} = not \ d efined$

$\displaystyle \lim_{x \to \infty} \frac{2x + 7 \sin x}{4x + 2 \cos x} \\ \\ \\ Take \ x \ as \ common \ on \\ both \ numerator \ and \ denominator} \\ \\ \\ \displaystyle \lim_{x \to \infty} \frac{x \left(2 + \dfrac{7 \sin x}{x} \right)}{x \left(4 + \dfrac{2 \cos x}{x} \right)}$

$\boxed{ \large { \displaystyle \lim_{x \to \infty} \frac{ \sin x}{x} = \displaystyle \lim_{x \to \infty} \frac{ \cos x}{x} = 0}}$

$\displaystyle \lim_{x \to \infty} \frac{7 \sin x}{x} = \displaystyle \lim_{x \to \infty} \frac{2 \cos x}{x} = 0$

$\dfrac{ \cancel{x} (2 +0)}{ \cancel{x }(4 + 0)} \implies \dfrac{2}{4} = \dfrac{1}{2}$

$\boxed {\bold{Hence \: \: \displaystyle \lim_{x \to \infty} \frac{2x + 7 \sin x}{4x + 2 \cos x} = \frac{1}{2} }}$

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