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

Question

Cameron Alexander · Accepted Answer

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#

Cameron Alexander · Answer

undefined To find the limit of [x + sin(x)] / [2x – 5sin(x)] as x approaches infinity, we can use the concept of limits. By dividing both the numerator and denominator by x, we get [1 + (sin(x)/x)] / [2 - (5sin(x)/x)]. As x approaches infinity, sin(x)/x approaches 0, so the limit simplifies to [1 + 0] / [2 - 0], which is equal to 1/2. Therefore, the limit of [x + sin(x)] / [2x – 5sin(x)] as x approaches infinity is 1/2.