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

##### 3 Answers

# 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

We can write this using asymptotic notation "

# 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

#### Explanation:

For

By the squeeze theorem,

Therefore

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

**Note**

For

Since

the squeeze theorem (at infinity) assures us that

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

Indefinite.

#### Explanation:

So, as

Applying L' Hospital rule. the limit is that for

The end x called

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

unspecific