solving limits by difinition

prove that the lim 1/x * sin(1/x) does not exist. (by the difinition)
Of course you do not mean exactly that.
limxa1xsin(1x)\displaystyle \displaystyle\lim _{x \to a} \frac{1}{x}\sin \left( {\frac{1}{x}} \right) does exist if a0\displaystyle a\ne 0.

So you must mean if a=0\displaystyle a=0 the limit does not exist.

Consider the sequence of points xn=2π(4n+1)\displaystyle x_n=\dfrac{2}{\pi(4n+1)}.
Can 1xnsin(1xn)\displaystyle \frac{1}{x_n}\sin \left( {\frac{1}{x_n}} \right) converge?
 
Show us what you have done so far.
Hi;
well... by controdiction method i suppose that this limit exists and it is L,
so by the difinition we have:
for every epsilon>0 there exist a delta>0 s.t !x!<delta ; !f(x)-L!<epsilon
by solving this we should face an counter...
 
Of course you do not mean exactly that.
limxa1xsin(1x)\displaystyle \displaystyle\lim _{x \to a} \frac{1}{x}\sin \left( {\frac{1}{x}} \right) does exist if a0\displaystyle a\ne 0.

So you must mean if a=0\displaystyle a=0 the limit does not exist.

Consider the sequence of points xn=2π(4n+1)\displaystyle x_n=\dfrac{2}{\pi(4n+1)}.
Can 1xnsin(1xn)\displaystyle \frac{1}{x_n}\sin \left( {\frac{1}{x_n}} \right) converge?
thank you pka, it helped me
 
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