# How do you find the limit of [x + sin(x)] / [2x – 5sin(x)] as x approaches infinity?

Mar 29, 2016

Piece by piece.

#### Explanation:

(x + sin(x)) / (2x – 5sin(x)) = x / (2x – 5sin(x)) + sin(x) / (2x – 5sin(x))

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First limit

lim_(xrarroo)x / (2x – 5sin(x)) = lim_(xrarroo) x/(x(2-5sinx/x))

$= {\lim}_{x \rightarrow \infty} \frac{1}{2 - 5 \sin \frac{x}{x}}$

Observe that ${\lim}_{x \rightarrow \infty} \sin \frac{x}{x} = 0$. (Prove it using the squeeze theorem.)

So the limit above becomes:

$= \frac{1}{2 - 5 \left(0\right)} = \frac{1}{2}$

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Second limit

lim_(xrarroo) sinx / (2x – 5sin(x))

Note that
${\lim}_{x \rightarrow \infty} 2 x - 5 \sin x = {\lim}_{x \rightarrow \infty} x \left(2 - 5 \sin \frac{x}{x}\right)$.

Now, since ${\lim}_{x \rightarrow \infty} \left(2 - 5 \sin \frac{x}{x}\right) = 2 - 5 \left(0\right) = 2$, we can conclude that

${\lim}_{x \rightarrow \infty} \left(2 x - 5 \sin x\right) = {\lim}_{x \rightarrow \infty} x \left(2 - 5 \sin \frac{x}{x}\right) = \infty$.

Therefore, we can again use the squeeze theorem to show that

lim_(xrarroo) sinx / (2x – 5sin(x)) = 0