Home > Calculus > Infinite Limits and Vertical Asymptotes

How do you find the limit of #(sin2x)/x# as x approaches infinity?

Answer 1

Emma Aldridge

#=0#

#sin 2x# is a continuous periodic function bounded as #sin 2x in [-1,1]#

therefore

#lim_(x to oo) (sin 2x)/x#

#= lim_(x to oo) c/x# with #c in [-1,1]#

#=0#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

To find the limit of (sin2x)/x as x approaches infinity, we can use the limit properties and trigonometric identities. By applying the limit definition, we have:

lim(x→∞) (sin2x)/x = lim(x→∞) sin(2x)/x

Using the Squeeze Theorem, we know that -1 ≤ sin(2x) ≤ 1 for all x. Therefore, we can write:

-1 ≤ sin(2x)/x ≤ 1/x

As x approaches infinity, the right side of the inequality, 1/x, approaches 0. Hence, we can conclude that:

lim(x→∞) (sin2x)/x = 0

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.