Question

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

Answer 1

`lim_(x rarr oo) (2x - sinx)/(3x+sinx) = 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 `2x` as the `-sinx` term becomes insignificant. Similarly the denominator behaves like `3x` 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 rarr oo) (2x - sinx)/(3x+sinx) = lim_(x rarr oo) (2x)/(3x)`
`" " = lim_(x rarr oo) 2/3`
`" " = 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.

Answer 2

`(2x-sinx)/(3x+sinx) = (2-sinx/x)/(3+sinx/x)` for `x != 0`

Explanation:

For `x != 0`, we have

`(2x-sinx)/(3x+sinx) = (2-sinx/x)/(3+sinx/x)`

By the squeeze theorem, `lim_(xrarroo)sinx/x = 0`. (See Note below.)

Therefore

`lim_(xrarroo)(2x-sinx)/(3x+sinx) = lim_(xrarroo)(2-sinx/x)/(3+sinx/x)`

`= (2-0)/(3+0) = 2/3`

Note

`-1 <= sinx <= 1` for all `x`.

For `x > 0`, we also have `1/x > 0`, so we can multiply to get

`-1/x <= sinx/x <= 1/x`.

Since `lim_(xrarroo)-1/x = lim_(xrarroo)1/x = 0`,

the squeeze theorem (at infinity) assures us that

`lim_(xrarroo)sinx/x = 0`.

Answer 3

Indefinite.

Explanation:

`|sin x|<=1`.

So, as `x to oo, (x-sin x)/(x+sin x) to` the indeterminate form

`oo/oo`.

Applying L' Hospital rule. the limit is that for `((2x-sin x)')/((3x+sin x)')`

`=lim (2- cos x)/(3+cos x)`

The end x called `oo` 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 `0/0 or oo/oo`, when a is the

unspecific `oo`. This is for the readers to ponder.