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

Question

William Abdalla · Accepted Answer

# lim_(x rarr oo) (2x - sinx)/(3x+sinx) = 2/3#

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.

Luke Alvarez · Answer

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

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#.

Samantha Adkison · Answer

Indefinite.

#|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.

Ezekiel Alexander · Answer

undefined The limit of (2x - sinx)/(3x+sinx) as x approaches infinity is 2/3.