How to prove limit as x approaches 0 of x sin(1/x)= 0 ?

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Oct 11, 2015

Answer:

#lim_(x rarr 0) xsin(1/x) = 0#

Explanation:

Here's an answer using the squeeze or sandwich theorem.

We know that any sine must be between -1 or 1, or in mathematical terms,

#-1 <= sin(theta) <= 1#

If #theta = 1/x# then

#-1 <= sin(1/x) <= 1#

Multiplying both sides by #x#

#-x <= xsin(1/x) <= x# for #x >0#
#-x >= xsin(1/x) >= x# for #x<0#

And since

#lim_(x rarr 0^+) x = lim_(x rarr 0^+) -x = 0#
#lim_(x rarr 0^-) x = lim_(x rarr 0^-) -x = 0#

We can use the squeeze theorem to say

#lim_(x rarr 0) xsin(1/x) = 0#

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Nov 28, 2016

Answer:

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

Explanation:

Here we are, one year later with a different way to solve this in case somebody comes across this:

#lim_(x->0)xsin(1/x) = lim_(x->0)sin(1/x)/(1/x)#

If we say #u = 1/x# then when #x -> 0#, #u -> oo# so we have

#lim_(u->oo)sin(u)/u #

Remember that the sine is always a number between -1 and 1, so when #u# grows without bound, the denominator becomes much too large and the whole fraction becomes closer and closer to zero. In other words:

#lim_(u->oo)sin(u)/u = 0#

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