Demonstration that: lim x→c f(x) = L ⇔ lim x→0 f(x+c) = L

[Moomin]

Im a first year student studying analysis and ODEs, and I cannot find this demonstration in my notes. This comes up in both modules and I would like to know the reasoning behind it.

Could you point me in the direction of a thorough proof online or help me here?

Thanks

 

[Ishuda]

Im a first year student studying analysis and ODEs, and I cannot find this demonstration in my notes. This comes up in both modules and I would like to know the reasoning behind it.

Could you point me in the direction of a thorough proof online or help me here?

Thanks

Let x = d + c or, equivalently d = x - c. Then as x goes to c, d goes to zero. So the first limit
\(\displaystyle \underset{x\, \to\,c}{lim}\, f(x) = L\)
becomes
\(\displaystyle \underset{d\, \to\,0}{lim}\, f(d\, +\,c) = L\)
But d is just a dummy variable and can be replaced with anything, including x.

 

[pka]

Im a first year student studying analysis and ODEs, and I cannot find this demonstration in my notes. This comes up in both modules and I would like to know the reasoning behind it.
Could you point me in the direction of a thorough proof online or help me here?

This is a thought problem. To say \(\displaystyle \displaystyle{\lim _{x \to c}}f(x) = L\) means:
if \(\displaystyle \forall x\ne c~\&~x\approx c\), x is 'close' to c, then \(\displaystyle f(x)\approx L\).

Now all the fancy \(\displaystyle \epsilon~\&~\delta \) terms are just used to formalize the definition of 'close'.

You must find a way to express this, \(\displaystyle x\approx 0~\iff~(x+c)\approx c\).