How do you find the limit of #sin((x-1)/(2+x^2)) # as x approaches infinity?

Answer 1

Emilia Aguilar

#lim_(x->oo)sin((x-1)/(2+x^2))=0#

Because #f(x)=sin(x)# is continuous, we have

#lim_(x->oo)sin((x-1)/(2+x^2)) = sin(lim_(x->oo)(x-1)/(2+x^2))#

#=sin(lim_(x->oo)(1/x-1/x^2)/(2/x^2+1))#

#=sin((0-0)/(0+1))#

#=sin(0)#

#=0#

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

Answer 2

Emily Ammerman

To find the limit of sin((x-1)/(2+x^2)) as x approaches infinity, we can use the squeeze theorem. By observing that -1 ≤ sin((x-1)/(2+x^2)) ≤ 1 for all x, we can conclude that the limit of sin((x-1)/(2+x^2)) as x approaches infinity is between -1 and 1. Therefore, the limit does not exist.

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.