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

3 Answers
Dec 16, 2016

# 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.

Dec 16, 2016

#(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#.

Dec 16, 2016

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.