#lim_(x->0)sin(1/x)/(sin(1/x))# ?

Find the limit #lim_(x->0)sin(1/x)/(sin(1/x))#

How would you approach this? Is it #1# or it doesn't exist?

2 Answers
May 24, 2018

# lim_(x rarr 0) \ sin(1/x)/(sin(1/x)) = 1 #

Explanation:

we seek:

# L = lim_(x rarr 0) \ sin(1/x)/(sin(1/x)) #

When we evaluate a limit we look at the behaviour of the function "near" the point, not necessarily the behaviour of the function "at" the point in question, thus as #x rarr 0#, at no point do we need to consider what happens at #x=0#, Thus we get the trivial result:

# L = lim_(x rarr 0) \ sin(1/x)/(sin(1/x)) #

# \ \ = lim_(x rarr 0) \ 1 #

# \ \ = 1 #

For clarity a graph of the function to visualise the behaviour around #x=0#
graph{sin(1/x)/sin(1/x) [-10, 10, -5, 5]}

It should be made clear that the function #y=sin(1/x)/sin(1/x)# is undefined at #x=0#

May 24, 2018

Please see below.

Explanation:

The definitions of limit of a function I use are equivalent to:

#lim_(xrarra)f(x) = L# if and only of For every positive #epsilon#, there is a positive #delta# such that for every #x#, if #0 < abs(x-a) < delta# then #abs(f(x) - L) < epsilon#

Because of the meaning of "#abs(f(x) - L) < epsilon#", this requires that for all #x# with #0 < abs(x-a) < delta#, #f(x)# is defined.

That is, for the required #delta#, all of #(a-delta,a+delta)# except possibly #a#, lies in the domain of #f#.

All of this the gets us:

#lim_(xrarra)f(x)# exists only if #f# is defined in some open interval containing #a#, except perhaps at #a#.

(#f# must be defined in some deleted open neighborhood of #a#)

Therefore, #lim_(xrarr0)sin(1/x)/sin(1/x)# does not exist.

A nearly trivial example

#f(x) = 1# for #x# an irrational real (undefined for rationals)

#lim_(xrarr0) f(x)# does not exist.