# Evaluate the limit  lim_(x rarr oo) (2x - sinx)/(3x+sinx)?

Dec 16, 2016

${\lim}_{x \rightarrow \infty} \frac{2 x - \sin x}{3 x + \sin x} = \frac{2}{3}$

#### Explanation:

This is not a vigorous proof. The sandwich theorem can be used if you need such a proof

$\sin x$ oscillates between $- 1$ and $+ 1$. as $x$ becomes large then the numerator behaves like $2 x$ as the $- \sin x$ term becomes insignificant. Similarly the denominator behaves like $3 x$ as the addition of $\sin x$ also becomes insignificant.

We can write this using asymptotic notation "~" meaning "behaves like" as follows;

 2x - sinx ~ 2x " as " x rarr oo
 3x + sinx ~ 3x " as " x rarr oo

And so for the quotient

(2x - sinx)/(3x+sinx) ~ (2x)/(3x) " as " x rarr oo

And so we can conclude that;

${\lim}_{x \rightarrow \infty} \frac{2 x - \sin x}{3 x + \sin x} = {\lim}_{x \rightarrow \infty} \frac{2 x}{3 x}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{2}{3}$
$\text{ } = \frac{2}{3}$

We can confirm this result graphically:
graph{(2x - sinx)/(3x+sinx) [-2, 30.04, -7.11, 8.91]}

Where you can see that for smaller values of $x$ the oscillation of the trig functions are apparent , but as $x$ increases the effect of these oscillations become infinitesimally small and the limit converges.

Dec 16, 2016

$\frac{2 x - \sin x}{3 x + \sin x} = \frac{2 - \sin \frac{x}{x}}{3 + \sin \frac{x}{x}}$ for $x \ne 0$

#### Explanation:

For $x \ne 0$, we have

$\frac{2 x - \sin x}{3 x + \sin x} = \frac{2 - \sin \frac{x}{x}}{3 + \sin \frac{x}{x}}$

By the squeeze theorem, ${\lim}_{x \rightarrow \infty} \sin \frac{x}{x} = 0$. (See Note below.)

Therefore

${\lim}_{x \rightarrow \infty} \frac{2 x - \sin x}{3 x + \sin x} = {\lim}_{x \rightarrow \infty} \frac{2 - \sin \frac{x}{x}}{3 + \sin \frac{x}{x}}$

$= \frac{2 - 0}{3 + 0} = \frac{2}{3}$

Note

$- 1 \le \sin x \le 1$ for all $x$.

For $x > 0$, we also have $\frac{1}{x} > 0$, so we can multiply to get

$- \frac{1}{x} \le \sin \frac{x}{x} \le \frac{1}{x}$.

Since ${\lim}_{x \rightarrow \infty} - \frac{1}{x} = {\lim}_{x \rightarrow \infty} \frac{1}{x} = 0$,

the squeeze theorem (at infinity) assures us that

${\lim}_{x \rightarrow \infty} \sin \frac{x}{x} = 0$.

Dec 16, 2016

Indefinite.

#### Explanation:

$| \sin x | \le 1$.

So, as $x \to \infty , \frac{x - \sin x}{x + \sin x} \to$ the indeterminate form

$\frac{\infty}{\infty}$.

Applying L' Hospital rule. the limit is that for $\frac{\left(2 x - \sin x\right) '}{\left(3 x + \sin x\right) '}$

$= \lim \frac{2 - \cos x}{3 + \cos x}$

The end x called $\infty$ is unspecific. So is end ( periodic ) cos x.

And so, the limit is indefinite.

Note: I am convinced that the limit is 2/3, from the different proofs,

including the one from Jim H. Yet, I would like to discuss the

possible failure of L'Hospital rule here, while applying the same on

$\lim x \to a$ of indeterminate forms $\frac{0}{0} \mathmr{and} \frac{\infty}{\infty}$, when a is the

unspecific $\infty$. This is for the readers to ponder.