delta-epsilon proof: if lim [x->a] f(x) = L, prove that

[maxx]

I'm confused about these type of problems:

Given lim x->a f(x)=L, prove that lim x->a [x+f(x)] = aL.

I can't assume the limit of the sum is the sum of their individual limits. I must use the delta-epsilon definition in my proof. HELP!

 

[daon]

Re: delta-epsilon proofs

I'm confused about these type of problems. Given lim x->a f(x)=L, prove that lim x->a [x+f(x)] = aL. I can't assume the limit of the sum is the sum of their individual limits. I must use the delta-epsilon definition in my proof. HELP!

This is not true. Lim x->a [x+f(x)] will exist so we can split the functions apart and it approaches a+L.

 

[pka]

From the given we have \(\displaystyle \L\left( {\varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left[ {0 < \left| {x - a} \right| < \delta \Rightarrow \;\left| {f(x) - L} \right| < \frac{\varepsilon }{2}} \right]\).
Now choose \(\displaystyle \delta ' = \min \left\{ {\delta ,\frac{\varepsilon }{2}} \right\}\) then follows that \(\displaystyle \L 0 < \left| {x - a} \right| < \delta ' \Rightarrow \;\left| {x + f(x) - \left( {a + L} \right)} \right| \le \left| {x - a} \right| + \left| {f(x) - L} \right| < \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon\).

 

[maxx]

cool thanks! I needed the delta epsilon proof.