How do you find the limit of #[x + sin(x)] / [2x – 5sin(x)]# as x approaches infinity?

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Answer 1 · 2016-03-29T16:15:17

Piece by piece.

#(x + sin(x)) / (2x – 5sin(x)) = x / (2x – 5sin(x)) + sin(x) / (2x – 5sin(x))#

.

First limit

#lim_(xrarroo)x / (2x – 5sin(x)) = lim_(xrarroo) x/(x(2-5sinx/x))#

# = lim_(xrarroo) 1/(2-5sinx/x)#

Observe that #lim_(xrarroo) sinx/x = 0#. (Prove it using the squeeze theorem.)

So the limit above becomes:

# = 1/(2-5(0)) = 1/2#

.

Second limit

#lim_(xrarroo) sinx / (2x – 5sin(x)) #

Note that
#lim_(xrarroo)2x-5sinx = lim_(xrarroo)x(2-5sinx/x)#.

Now, since #lim_(xrarroo)(2-5sinx/x) = 2-5(0) = 2#, we can conclude that

#lim_(xrarroo)(2x-5sinx) = lim_(xrarroo)x(2-5sinx/x) = oo#.

Therefore, we can again use the squeeze theorem to show that

#lim_(xrarroo) sinx / (2x – 5sin(x)) = 0#