ANother delta-epsilon proof: prove xf(x)->aL

[maxx]

I would appreciate the help!

Given that lim x->a f(x)=L, prove that lim x->a [x*f(x)]=aL. Use the delta epsilon definition and you may use that the lim x->a of [x+f(x)]=a+L.

Thanks in advance! you all were very helpful last time.

 

[pka]

Here are the basic steps.
\(\displaystyle \begin{array}{rcl}
\left| {aL - xf(x)} \right| & = & \left| {aL - af(x) + af(x) - xf(x)} \right| \\
& \le & \left| {aL - af(x)} \right| + \left| {af(x) - xf(x)} \right| \\
& \le & \left| a \right|\left| {L - f(x)} \right| + \left| {f(x)} \right|\left| {a - x} \right| \\
\end{array}\)

Now for x near a, f(x) is bounded and near L.