How do you determine the limit of #sin(x)/x# as x approaches infinity?

Question

Emma Aldridge · Accepted Answer

#lim_(x->oo)sinx/x=0#

We'll use the Squeeze Theorem, which states that if we have some function #f(x),# we define new functions #h(x), g(x)# such that

#h(x)<=f(x)<=g(x)#

Then, we'll take #lim_(x->a)h(x), lim_(x->a)g(x).#

If these limits are equal, then #lim_(x->a)f(x)=lim_(x->a)h(x)=lim_(x->a)g(x)#

Or, all the limits are equal.

So, recall that

#-1<=sinx<=1#

Then,

#-1/x<=sinx/x<=1/x#

Take #lim_(x->oo)-1/x, lim_(x->oo)1/x:#

#lim_(x->oo)-1/x=-1/oo=0#

#lim_(x->oo)1/x=1/oo=0#

So,

#lim_(x->oo)sinx/x=0#

Elijah Abram · Answer

undefined To determine the limit of sin(x)/x as x approaches infinity, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate both the numerator and denominator with respect to x. The derivative of sin(x) is cos(x), and the derivative of x is 1.

Taking the limit of the derivative of sin(x) divided by the derivative of x as x approaches infinity, we get the limit of cos(x)/1. Since the cosine function oscillates between -1 and 1, there is no definite limit for cos(x) as x approaches infinity. Therefore, the limit of sin(x)/x as x approaches infinity is undefined.