solving limits by difinition

[fatemeh8989]

prove that the lim 1/x * sin(1/x) does not exist. (by the difinition)
thank you!

 

[srmichael]

Show us what you have done so far.

 

[pka]

prove that the lim 1/x * sin(1/x) does not exist. (by the difinition)

Of course you do not mean exactly that.
\(\displaystyle \displaystyle\lim _{x \to a} \frac{1}{x}\sin \left( {\frac{1}{x}} \right)\) does exist if \(\displaystyle a\ne 0\).

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

Consider the sequence of points \(\displaystyle x_n=\dfrac{2}{\pi(4n+1)}\).
Can \(\displaystyle \frac{1}{x_n}\sin \left( {\frac{1}{x_n}} \right)\) converge?

 

[fatemeh8989]

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...

 

[fatemeh8989]

Of course you do not mean exactly that.
\(\displaystyle \displaystyle\lim _{x \to a} \frac{1}{x}\sin \left( {\frac{1}{x}} \right)\) does exist if \(\displaystyle a\ne 0\).

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

Consider the sequence of points \(\displaystyle x_n=\dfrac{2}{\pi(4n+1)}\).
Can \(\displaystyle \frac{1}{x_n}\sin \left( {\frac{1}{x_n}} \right)\) converge?

thank you pka, it helped me